Review – Dublin’s Cathedrals

Ireland — the “Land of Saints and Scholars” on many a tourist brochure — has no shortage of religious attractions for churchy types and Sunday-morning snoozers alike. With 1600 years of Christian tradition, and who knows how many years of pre-Christian religious practice before that, it’s difficult to throw a rock from the top of one of Ireland’s many green hills without knocking a holy well, a grave dolmen, a ruined abbey, or a lady with a big hat walking out of Sunday service.

Dublin has a number of churches worth a visit — consider a church-inside-a-church at St. Audeon’s or a visit to St. Valentine’s remains at Whitefriar Street — but the holy hits of the capital are the two no-longer-Catholic cathedrals of Christ Church and Saint Patrick’s.

Due to some complicated and painful history, these two massive cathedrals sit no more than a few hundred yards from one another, and both are of the Church of Ireland — a not-English-thank-you-very-much cousin of the Church of England. Every year, thousands of hopeful Irish-American Catholics plan to attend Mass at one of these cathedrals — especially St. Paddy’s — and are confused by the raised pulpit, the stodgy old hymns, and those comically massive ladies’ hats.

Christ Church Cathedral

Closest to Dublin’s city centre, and probably the first one you’ll see, the blocky steeple of Christ Church stands tall in the heart of the city’s medieval quarter. The first church here would have stood within Dublin’s original stone city walls (sections of which are still on display in the surrounding neighborhood), but today, it’s surrounded by the ugly 1980s-era Wood Quay Civic Offices and busy streets.

Christ Church Cathedral

In the courtyard, take a look at the church from different angles, checking out the ruins of the old chapter house. The sharp corners and square towers are quite photogenic — especially in the early morning and at sunset.

Inside, explore the large main sanctuary (technically free if you are really going in just to pray and not to sightsee) and try to find the notable graves, particularly the stone sarcophagus of Strongbow — the English lord who brought the Normans to Ireland and a tasty brand of hard apple cider to the rest of the world.

The real star of the Christ Church show is downstairs in the crypt. Most of the burial chambers have been converted to museum showpieces for permanent and temporary displays about the history of the cathedral and Viking-Age Dublin. The most famous posthumous residents of the crypt — maybe even more than ol’ Strongbow himself — are the mummified cat and rat found in an old organ pipe, naturally preserved by the dry conditions underground.

Saint Patrick’s Cathedral

Reportedly built on the site of a holy well in which Saint Patrick is said to have baptized the local Pagan king (accidentally stabbing the hapless king through the foot with his staff in the proceedings…ouch). Lending credence — but not really any hard evidence — to the legend is the fact that the River Poddle, which has long since gone underground, flows near where the church stands today.

Saint Patrick’s Cathedral

Nothing remains of the earliest structures of the church, and most of what you can see today is a smattering of bits, bobs, patches, and restorations going back a thousand years or so. After the church was seized by the Anglicans during the Reformation — well, at least Henry VIII got his divorces — it suffered a period of decline and it was the Guinness family, yes, that Guinness family, who stepped in to help fix the place up.

On your visit today, you’ll see a well-presented church with a wealth of details to be enjoyed by the sharp-eyed tourist. Look on the floor for the grave of Jonathan Swift, author of Gulliver’s Travels and once the dean of the cathedral; find the dedication to one Mr. Wallop, who took a bad military defeat and from whom we get the term “walloped.” Those who have been paying attention to Irish history will notice the chair once occupied by one William of Orange to celebrate his victory of the Irish Catholics at the pivotal Battle of the Boyne.

If you can swing it, time your visit with one of the daily guided tours given by knowledgeable guides — especially if you don’t know off-hand who William of Orange was. The guided tour visits the most interesting corners of the massive cathedral, and the guides clarify many confusing points: indicating which chairs are reserved for the President of Ireland and which are reserved for the Prime Minister, and what the heck is the difference between the two; and pointing out the flags of notable wealthy, landed families whose flags still hang over the central choir, and what the heck the obscure heraldry means.

If you can’t make the guided tour and are still determined to visit, maximize your admission fee by grabbing one of the audio tours and take your time with it. Without some kind of interpretive material, you’ll find yourself a bit overwhelmed in the dimly lit cavern that is the St. Patrick’s sanctuary.


Meningitis limb loss boy told to prove disability Published duration 19 February 2016

image caption Edward Bright lost all his limbs when he contracted meningitis as a young boy

The family of a teenager who lost his arms and legs from meningitis were shocked when told he had to prove he still needed a disability payment.

Edward Bright, from Ripley, Derbyshire, lost his limbs when he was seven and needs help with much of his life.

When he turned 16 and reapplied for a benefit he was told he had to attend a meeting to be assessed.

The Department for Work and Pensions has now said a face-to-face meeting is not necessary.

image copyright Bright family image caption Edward Bright as a young boy in hospital with meningitis

Edward's family claimed disability living allowance (DLA) on his behalf to help towards the added cost of life as a quad-amputee but when he turned 16 he had to reapply for a personal independence payment (PIP).

The family were horrified to receive a letter calling him to a meeting with an assessor in Derby and threatening to withdraw the money if he did not attend.

His dad Steve said: "He has already been through all the assessments. Nothing has changed [in his situation] in the last seven years; nothing will change, so being called in to interviews every year is pointless.

image caption Edward Bright was sent a letter saying he needed to meet assessors face-to-face

"People with disabilities have enough appointments anyway. You should reach a level when it's flagged up on the system: no more interviews needed."

Since the family's case featured in the Daily Mirror , the Department of Work and Pensions has admitted the assessment is unnecessary.

"All DLA claimants are invited to apply for the working-age benefit Personal Independence Payment when they turn 16. In this instance a paper-based assessment has been completed and there is no need for a face-to-face assessment.

"We've tried to call the Brights numerous times to make clear there's no need to attend a meeting on Monday. A letter has been sent through the post."


Q:

How to read a large file line by line?

I want to read a file line by line, but without completely loading it in memory.
My file is too large to open in memory, and if try to do so I always get out of memory errors.
The file size is 1 GB.

A:

You can use the fgets() function to read the file line by line:
$handle = fopen("inputfile.txt", "r");
if ($handle) {
    while (($line = fgets($handle)) !== false) {
        // process the line read.
    }

    fclose($handle);
} else {
    // error opening the file.
}

A:

if ($file = fopen("file.txt", "r")) {
    while(!feof($file)) {
        $line = fgets($file);
        # do same stuff with the $line
    }
    fclose($file);
}

A:

You can use an object oriented interface class  for a file - SplFileObject http://php.net/manual/en/splfileobject.fgets.php (PHP 5 >= 5.1.0)
<?php

$file = new SplFileObject("file.txt");

// Loop until we reach the end of the file.
while (!$file->eof()) {
    // Echo one line from the file.
    echo $file->fgets();
}

// Unset the file to call __destruct(), closing the file handle.
$file = null;


Q:

Операторы Lua в C# операторы (необычное сравнение)

Никак не могу понять как перевести этот оператор из Lua в C# оператор?
--Это некая карта из массива байт
--0x0 используется как ничего, а 0x1 как стена
local map = {
{ 0x1, 0x1, 0x1, 0x1 },
{ 0x1, nil, nil, 0x1 },
{ 0x1, nil, 0x1, 0x1 },
{ 0x1, 0x1, 0x1, 0x1 },
}

local yMap, xMap = math.floor(player.y / 32), math.floor(player.x / 32)

if map[yMap] and map[yMap][xMap] then
    return map[yMap][xMap]
end

Ну к сути - Я никак не понимаю как это вообще работает в Lua?
Я про строчку с сравнением: if map[yMap] and map[yMap][xMap] then
Я пытался понять и написать это на C#
if ((map[yMap] != 0x0) & (map[yMap][xMap] != 0x0))

Но все тщетно, компилятор конечно же ругается что я сравниваю byte[] и byte.
Никак не пойму почему на Lua это работает?! Может он берет байт от начала массива?
И да, как я правильно понял:
if map[0] then и if map[0] not nil then это одно и тоже?
То есть Если НЕ nil вернет true?
-
В Lua почти ничего не знаю, просто нужно Lua скрипт в C# перенести, пожалуйста, объясните глупцу, как это работает?

A:

if map[yMap] - да это тоже самое что спросить map[yMap]  существует?
В строго типизированных языках двумерный массив определен заранее, как у вас. И эту проверку можно опустить.
      public struct Coord    { public int x, y;  }

           Coord player;
        player.y = 3992;
        player.x = 62;

        int[,] map = {
            { 0x1, 0x1, 0x1, 0x1 },
            { 0x1, 0x0 , 0x0, 0x1 },
            { 0x1, 0x0 , 0x1, 0x1 },
            { 0x1, 0x1, 0x1, 0x1 },
            { 0x1, 0x1, 0x1, 0x1 },
        } ;

        int xMap = (int)Math.Floor( (double) player.x / 32);
        int yMap = (int)Math.Floor( (double) player.y / 32);

        Console.WriteLine(  );

        if (xMap<=map.GetLength(1) & yMap<=map.GetLength(0))
            if  (map[yMap,xMap] == 0x0)
              { Console.WriteLine("Ok"); }
            else  Console.WriteLine("None");
        else  Console.WriteLine("Bad index");
    }

Но если есть необходимость делать проверку, то как-то так:
(map.GetType().IsArray) && (map.Rank==2)


ose -4*r + 47 = 4*i + 19, 5*i + b = 4*r. Let u = -6 + r. Solve -3*g = u*g - 20 for g.
4
Let a = -74 + 74. Solve -2*v = -a + 2 for v.
-1
Suppose 60 = 4*u + 4*h, 0*h = 4*h + 20. Solve -u = -c - 3*c for c.
5
Let p(x) = 39*x + 7 - 41*x + 1. Let o be p(4). Solve 0 = -o*d - d for d.
0
Suppose -6*o + 2*o = -32. Suppose 4*r + o = -3*g, 2*r - 3 = 3*g - 25. Suppose -g*z + 9 = y + z, 0 = 5*z - 5. Solve n - y*n = 0 for n.
0
Suppose 0 = 3*n - 6*n + 15. Let a = 7 - n. Solve 0 = -a*q - 0*q - 8 for q.
-4
Suppose 3*x = y - 6, -4*y = 4*x - 4 - 4. Solve -15 = y*i - 0*i for i.
-5
Let x = 36 - 34. Solve x = -w - 0 for w.
-2
Let y = 14 + -8. Suppose -k - 3 = 0, 5*p + 10 = -2*k - y. Let m be (-1 - -3) + p - 0. Solve 2*n + n = m for n.
0
Let u be 12/18*(-63)/(-6). Solve -9 = 4*z - u*z for z.
3
Let x(y) be the second derivative of y**5/20 + y**4/4 + y**3/3 + y**2 - 2*y. Let s be x(-2). Let o be s/4 - 9/(-2). Solve -4*v + o = -5*v for v.
-5
Let v = 19 + -17. Solve -u = -v*u - 5 for u.
-5
Let p(c) = c**3 - 5*c**2 + 2*c + 9. Let a be p(4). Let q be (2 - a)*(0 + 4). Solve 2 - 14 = q*s for s.
-3
Let z be ((-1)/((-6)/(-10)))/(9/(-27)). Solve 4 = z*i + 14 for i.
-2
Let z be (-24)/(-9)*9/6 + 3. Solve -2*x + 3 = -z for x.
5
Let m = 10 - 4. Let i(f) = 8*f - 3. Let u be i(m). Suppose -4*t = -7*t + u. Solve -3*g - t = 2*g for g.
-3
Let u(b) = 2*b**2 + 2*b - 2. Let j be u(2). Suppose 0*q = -5*q + j. Suppose -p = -q*p. Solve p = -0*k + 2*k + 6 for k.
-3
Let n be -12*(-2)/(-8)*2. Let f be (n/(-4))/(1/2). Let r be (-1)/5 - f/(-15). Solve m = -r*m for m.
0
Let y be 9/12 - 27/(-12). Suppose 5*v = y*v + 22. Solve v = -4*j + 3 for j.
-2
Let q(g) = g + 3. Let z be q(-3). Suppose z = -2*r + r. Solve -3 = -f - r*f for f.
3
Suppose -4*a - 5*z + 45 = 0, -a = a + 4*z - 30. Solve 3*u - 14 + a = 0 for u.
3
Let j(f) = f**2 - f + 4. Let g be j(0). Let s be g - 4*(-1)/(-2). Solve n = s*n for n.
0
Suppose -4*w = 4*l + 4, 0 = 3*w - 3*l - 6 - 21. Suppose -2*c - w = 2, 2*v - 5*c - 15 = 0. Solve v = 2*z + 2*z - 12 for z.
3
Suppose -3*b = 5*s - 16, 4*b + 0*s - 12 = -2*s. Solve -b*u = u for u.
0
Suppose 0 = -5*j + 5*s + 35, 4*s = 2*j - s - 20. Suppose 0 = -3*l + j*c + 29, 0 = -l - 2*l + 4*c + 25. Solve -r = l*r for r.
0
Let n(u) = -2*u**3 + 19*u**2 - 17*u - 3. Let x(r) = -r**3 + 10*r**2 - 9*r - 1. Let t(s) = -4*n(s) + 7*x(s). Let p be t(5). Solve 0*h - p = h for h.
-5
Let s be 1*(1 - (-2)/(-1)). Let g = s - -3. Suppose 270*q = 275*q - 20. Solve -10 - g = -q*l for l.
3
Let v be 3*(-1)/(-3) - 12. Let u = 17 + v. Solve 6*l = 4*l - u for l.
-3
Suppose 2*u = -u + 12. Suppose 4*n + h = 9, -4 = -5*n - u*h - 1. Solve -4 = -n*t - t for t.
1
Let r = -107 + 111. Suppose -15 = -5*o - 3*u, 2*u - 1 = o + 9. Solve o = -r*v + 2 + 10 for v.
3
Let y be ((-7)/(-2))/((-12)/(-24)). Let j(a) = a**3 - 7*a**2 + 2*a - 9. Let l be j(y). Solve -2*g + 12 = -l*g for g.
-4
Let j be 3*(-3 + 44/12). Solve -5*u = -j - 3 for u.
1
Suppose 0*a + 5*a = 0. Solve -2*z + 2 + 4 = a for z.
3
Suppose -5*r + 4*i = -3 - 13, r = -5*i - 20. Suppose -n + 1 + 2 = r. Solve -3*p - n = -0 for p.
-1
Let c(o) be the third derivative of -o**6/120 + o**5/15 + o**3/3 + o**2. Let v be c(4). Solve -2*j - 2 - v = 0 for j.
-2
Let n be 2/3*(-39)/(-2). Suppose -5*j - 5*d + d = 83, d = j + n. Let f be ((-9)/j)/((-1)/(-5)). Solve 0 = -f*s - 8 - 7 for s.
-5
Suppose -3*p + 0 + 12 = 0. Solve p = -b - 3*b for b.
-1
Let u be 1 + -1 + 6/2. Suppose u*g = -q + 18, -5*q = -g - 4*g - 70. Solve 0 = -2*l - 3*l + q for l.
3
Let r = -10 + 9. Let f be (r - -1)*(2 - 3). Suppose 3*l - l - 10 = f. Solve 0 = -l*c + 8 + 2 for c.
2
Let h = -6 + 11. Let v be 4*(0 + 3 + -1). Solve -33 = h*k - v for k.
-5
Let n be 2 + 1 + -1 + 2. Suppose -1 + 17 = n*y. Suppose -y*a + 9 = -3. Solve -a*m = -m + 6 for m.
-3
Let c = 15 + -13. Solve 2 = -c*w + 8 for w.
3
Suppose 33 = 6*u - 15. Solve -4*r + u = -2*r for r.
4
Suppose 5*n + 9 + 11 = 0. Let r be (-3)/12 - 49/n. Let p = -10 + r. Solve 3 = l + p*l for l.
1
Suppose l - 5*l + 52 = 0. Suppose 4*h = l - 1. Solve h*r + 1 + 2 = 0 for r.
-1
Let n = 5 + 1. Suppose -v - 3*b = b + n, 33 = 3*v - 5*b. Solve 0 = 5*d - 3*d - v for d.
3
Let t(v) = v**2 - 7*v + 2. Let q be t(7). Solve q*o = -0*o - 4 for o.
-2
Suppose t - 8 = -3*t. Solve 7*z - 20 = t*z for z.
4
Let n(k) = -k**2 + 6*k + 8. Let x be n(6). Suppose -5*z + 17 = -3*b - 3, 0 = 2*z - 2*b - x. Solve 4*g - 4 = z for g.
2
Let f(r) = r**3 - 3*r**2 + r - 1. Let o be f(3). Suppose 3*p = 2*p + 5*z + 28, -4*p = -z - 17. Solve p*q - o = 2*q for q.
2
Let o be -2 - (2*-1 - 10). Suppose -2 = 4*t - 2*g, 0*t - t + 2*g - 8 = 0. Suppose -3*x - 5 = -n, t*n + 4*x + x + 1 = 0. Solve n*j + o = 2 for j.
-4
Let p = -10 - -14. Suppose 4*v - 8 = 0, 0*v = -3*k + p*v - 5. Suppose -a - 18 = 4*i, -a + 2*a + 8 = -2*i. Solve 3*m + a + k = 0 for m.
-1
Suppose 17 = 5*v + 2. Let o be -10*(12/(-10))/v. Solve -2*g + o - 12 = 0 for g.
-4
Suppose 0 = 2*m - 1 - 33. Let q = 17 - m. Let v = 15 - 9. Solve q = w - v*w for w.
0
Let g = 9 + -7. Suppose g*n = -0*n + 64. Suppose 2*t = 4*k - 18, 0 = 3*k + 3*t + 5 - n. Solve -4*m = -k*m + 2 for m.
1
Suppose 0 = -3*q + 5 + 1. Let n be (q*-5)/(-2) - 2. Let c be 2/(-11) + (-1027)/(-143). Solve -1 = -n*v - c for v.
-2
Let y = 51 - 50. Solve o + 6 = y for o.
-5
Suppose 3 + 2 = -t, o + 5*t = -18. Let p(u) = -u + 8. Let x be p(o). Solve 5*s - x = 9 for s.
2
Let m(z) = 4*z + 1. Let w be m(0). Solve -5 = -2*r + w for r.
3
Let n(j) = j**3 + 9*j**2 - 11*j - 2. Let y be n(-10). Solve y = -3*f - f for f.
-2
Let u(z) = -z**3 - z**2 - 1. Let b be u(-2). Suppose 0 = -5*d + a + 3*a + 22, 3*d + 9 = -5*a. Suppose h = 4*i, 5*i = b*h - d*h + 1. Solve -4*o + h = -3*o for o.
4
Let k(i) = 14*i**2 - 3*i - 3. Let f be k(-1). Solve -f = 3*s + 1 for s.
-5
Let o be (-4)/((-24)/9)*6. Let l = 9 - o. Solve -3*z + 9 = -l for z.
3
Suppose -q + 5*r = -6*q + 95, -4*r = -3*q + 29. Solve 4*n = 7*n + q for n.
-5
Suppose -3*k + 37 = -5*r, 5*k - 5*r - 33 = 3*k. Solve -k*c = -14 - 6 for c.
5
Let k = -14 - -25. Solve 2*a + k = 1 for a.
-5
Let o = -3 - 0. Let b be -4*(9/4)/o. Solve -m - b*m = -4 for m.
1
Let g = 4 + -2. Suppose -4*s + 2 = -g*s. Suppose -4*c - s + 17 = 0. Solve -w + 0 = -c for w.
4
Let s be (12/(-8) - -1)*-22. Solve 2*b + s = 3 for b.
-4
Let p be -15 - (-2 - 2/(-1)). Let h be (0 + (-18)/p)*5. Let s(j) = -j**2 - 5*j - 4. Let l be s(-4). Solve 5*w - 3*w - h = l for w.
3
Let c(b) = -b - 14. Let a be c(6). Let o = -20 - a. Solve -1 = -l - o*l for l.
1
Suppose 3*z = -4*v + 15, 4*v - 27 = 2*z - z. Let x = v + -3. Let u = -4 + 16. Solve -x*r + 6*r = u for r.
4
Suppose 60 = -0*x + 5*x. Suppose 0 = -3*i + 3*r + x, i + 5*r = 2*i. Solve -20 - i = 5*v for v.
-5
Let w(y) = -y + 2. Let k(m) = m**2 + 2*m - 3. Let q be k(-3). Let b be w(q). Suppose -2*a = 2*a - 2*s - 10, 3 = b*a - 3*s. Solve -2*g = -1 - a for g.
2
Let x = -119 + 119. Let i(c) = -c**3 - 2*c**2 - c + 1. Let s be i(-2). Solve -2*o - s - 1 = x for o.
-2
Suppose -13 = -a - 2*l + 12, 0 = -3*l + 9. Solve 5*u = -a + 44 for u.
5
Suppose 5*w = -0*w - 70. Let v be w/(-8) + (-4)/(-16). Solve v*z = 7*z + 5 for z.
-1
Let a = -108 - -108. Solve a = 5*n - n for n.
0
Let w = 28 - 24. Suppose -4*r + 6 = -2*r + w*j, -3*r = -2*j - 9. Solve 2*z + r*z + 10 = 0 for z.
-2
Suppose 3*a = -5*c + 26, 0*a + 4*c = -a + 18. Solve r + a*r + 6 = 0 for r.
-2
Suppose -35 = -5*v - 10. Solve n - 20 = v*n for n.
-5
Suppose 0 = -s + 4, 5*s - s - 112 = -2*w. Suppose 3*n + 4*m - w = 0, 2*m = -2*n + 4*n - 18. Let k = n + -7. Solve 0 = 3*t - 2*t - k for t.
5
Suppose -4*r + 6 = 3*q, -4*r - 2*q + 4 = r. Let z be 0 - 1/(1/(-5)). Suppose -33 - 27 = -z*i. Solve r*j = -3*j - i for j.
-4
Let q(x) = 3*x**2 - x + 2. Let w be q(2). Solve 0*p = 4*p + w for p.
-3
Let a be 2/5*(2 - -8). Suppose 2*l = -0*l + 5*g - 14, a*l + 28 = 2*g. Let m be l/(-2) + 2/(-4). Solve 6 = m*f - 3 for f.
3
Let z(w) = -w**2 + 7*w + 2. Let r be z(7). Solve -u = -r*u for u.
0
Let w(z) = -z**3 + 3*z**2 - 3*z + 3. Let f be w(3). Let p be (-22)/f - 2/(-6). Solve -p = -m - 0 for m.
4
Suppose -y = 3*n - 5, -4*y + 8 = -n + 1. Solve 0*u = u + y for u.
-2
Suppose 0 = -l - 0 + 25. Let u(x) = -x + 10. Let j be u(8). Suppose -5*i = -3*c + l, -j*c + 2*i = c - 10. Solve -3*g + g = c for g.
0
Let c(r) = -3*r + 2. Let n be c(-3). Suppose 2*x = 3 + n. Solve -5*p = 8 + x for


and 12/(-18) - (-1 - 118/36).
90
Suppose -36 = -3*j + 3*k, 0*j + j - 5*k - 12 = 0. Suppose -4*g - 1 + 65 = 0. Calculate the smallest common multiple of g and j.
48
Suppose 2*d - d - 9 = z, 81 = 5*d + 4*z. Suppose 6*v - 18 = 3*v. What is the smallest common multiple of v and d?
78
Find the common denominator of 84*(-1)/(16/(-6)) and -29/18.
18
Let y be (-195)/(-12) + (-2)/(-8). Suppose 3*b - 1 + 4 = 0. Let t = y + b. Find the common denominator of t and 35/16.
16
Suppose 2*y - 11 = 3. Suppose d + d - 6 = 0. Suppose 0 = c - 2*c + d. What is the least common multiple of y and c?
21
Calculate the common denominator of 4/18 + 6377/(-630) and -11/18.
90
Suppose -4*w = 9 - 1, -w + 18 = 2*v. What is the least common multiple of 4 and v?
20
Let q(j) = j**2 + 2*j - 4. Let x = 16 - 13. Suppose k = -0 + x. Calculate the least common multiple of q(-5) and k.
33
Let j be 3/4 + 1092/(-48). Let d = j + 24. Suppose -4*p - 1 = -b + 4, 0 = 4*b - 4*p - 8. Calculate the smallest common multiple of d and b.
2
Suppose -83*z = -84*z + 14. Calculate the least common multiple of z and 12.
84
Suppose -2*u = -h - 319, -68 - 88 = -u + 4*h. Calculate the common denominator of -17/14 and u/(-6) - 4/(-24).
14
Let l = 18 + 10. What is the lowest common multiple of 2 and l?
28
Let z = 100 + -76. What is the lowest common multiple of 36 and z?
72
Find the common denominator of 5/90*357 - 4 and ((-40)/6)/(3/2).
18
Let m(h) = -h**2 + 6*h + 6. Let f be (1 - -1) + -3 + 24. Suppose 2*w = f + 15. Find the common denominator of w and m(7).
1
Let h(v) = 4*v - 1. Let c be h(1). Suppose -c*k = 3, -d + 20 = d - 4*k. Calculate the least common multiple of d and 10.
40
Let c be -2*1/3*3. Let p(m) = -m - 2. Let r be p(4). What is the lowest common multiple of 3 and (r/(-1) - c) + 0?
24
Let a be (0 + 1)/((-1)/3613). Let y be (2 - -5)*a + 0. Let z = 556299/22 + y. Find the common denominator of z and 91/20.
220
Let o be 2/1 - (1 + -7). Let f(m) = 11*m - 4*m**3 - 2*m + 5*m**3 - 9*m**2 - 2. Calculate the least common multiple of f(o) and 1.
6
Suppose -9*c + 260 = -100. What is the smallest common multiple of 8 and c?
40
Calculate the lowest common multiple of 2 and 1/(-2 - 0 - (-55)/22).
2
Suppose 0 = -5*h + 15 + 140. Suppose 5*j - 14 - h = 0. What is the common denominator of (146/j + 0)/(-4) and 95/8?
72
Let o = -91/22 - -46/11. Let v = -2264/21 - -320/3. Let t = 15/14 + v. What is the common denominator of o and t?
154
Calculate the common denominator of 18/11 and (4 - 1) + 850/(-51).
33
Suppose 3*j - 6 = 3*r, -5*j - 5*r = -2*r - 10. Suppose -4 = 5*i + 1. Let o = 3 + i. What is the least common multiple of o and j?
2
Let d = -4/1741 - 231529/10446. Suppose 2*f + 20 = 4*g, 3*f = 8*f - 5*g + 25. What is the common denominator of d and -4 - 350/(-12) - f?
6
Let w = 42012831 - 51255651039/1220. Let i = w - -5/244. What is the common denominator of i and 103/14?
70
Let j = 2159/4 - 561. Let c be (-600666)/164000 + 4/(-10). Let t = -1/10250 - c. Calculate the common denominator of t and j.
16
Suppose 0 = -3*h + 6*h + 3*m + 3, -4*h = -2*m - 20. Suppose -h*w - 6 = -6*w. Calculate the common denominator of (w/(-70))/((-6)/1455) and -71/16.
112
Suppose -5*o + 3*i = -0*o - 29, 0 = 5*i + 15. What is the smallest common multiple of 12 and o?
12
Suppose 4*z - 20 = -0*z. What is the least common multiple of z and 4?
20
Let k = 1/142 + -3841/994. Calculate the common denominator of 65/4 and k.
28
Let w(p) = -p**3 - 6*p**2 - 6*p - 5. Let s be w(-5). Suppose -3*m + 15 = -s*m. Let q = 6 - 1. Calculate the lowest common multiple of m and q.
5
Suppose -6*d = -4*d + 52. Calculate the common denominator of d and 13/4.
4
Let t = 10 + -5. Suppose -t*c = -48 - 52. Calculate the least common multiple of ((-9)/(-3) + 1)/1 and c.
20
Let m = 31 + -12. Suppose 0 = 2*h - 6 - 0. Let d = h + m. What is the least common multiple of d and 6?
66
Let r be 2/(-5) + 60/25. Suppose 0 = -l - 2*b + 8, r*b = 4*l - 0*b - 12. Let p(n) = -9*n + 1. What is the least common multiple of p(-1) and l?
20
Let q be 2*(-1 - (-34)/2). Suppose 2*p - 6*p = -q. What is the least common multiple of p and 3?
24
Let a = -4 + 5. Calculate the lowest common multiple of 22 and 3*a + (14 - -3).
220
Let l be (-213)/(-6)*(-26 - 1). Let t = l - -990. Calculate the common denominator of (-65)/16 + 2 + 1 and t.
16
Suppose -3*k = -3 - 9. Calculate the common denominator of 111/(-108) + k/16 and 45/2.
18
Let l = 8180/11 - 741. Let p be -2 - -1498*(-23)/11. Let g = 3130 + p. Find the common denominator of g and l.
11
Let l = -34 + 44. Calculate the least common multiple of 16 and l.
80
Let j = -474604827852440507/68342094980908326 - 1/385891153013. Let h = 67042933/127907 + -6814/13. Let r = j - h. Find the common denominator of r and -85/12.
36
Calculate the common denominator of (-18)/72 + (-26)/8*2 and -81/26.
52
Suppose 3*b + 2*j = 20, -2*b + 8 = 3*b - 3*j. Suppose 3*w + 22 = b*w. What is the smallest common multiple of w and 8?
88
Suppose -l = -2*w - 82, 0 = -2*l - w + 6*w + 161. Calculate the smallest common multiple of l and 20.
440
Let l = 11 + 0. Suppose -3*s + 13 + l = 0. Calculate the least common multiple of 4 and s.
8
Let n = -1 - -4. Let u = -8 + n. Find the common denominator of -46/11 and (1/22)/(u/(-230)).
11
Let g be (-2)/(-8) - 2479/(-68). Let m = g - 2843/102. Let h = -1530/59 - -24523/1180. What is the common denominator of h and m?
60
Suppose 3*z + z = -2*h + 38, -32 = -3*z - 5*h. What is the lowest common multiple of 9 and z?
9
Suppose i = 5*o - 2 - 14, -5*i = -4*o + 17. Find the common denominator of 81/10 and o - (2 - (-297)/24).
40
Let a = 15 - 9. Let c be (-4)/(-3)*a/(-4). Let g = c - -10. What is the least common multiple of 10 and g?
40
Let v = 16 - 10. What is the smallest common multiple of v and (-276)/(-27) + 2/(-9)?
30
Let g = 2231/2035 + -1/185. Let k = -17495/7 + 2491. Find the common denominator of g and k.
77
Calculate the common denominator of 23/10 and (-1 - -1) + 236/16.
20
Let s(u) = u**3 - 5*u**2 - 2*u + 7. Let h be s(5). What is the common denominator of 59/3 and -47 + 3*(-2 - h)?
3
Let a = -581 - -6386/11. Let w = -23057/10 + 2308. Find the common denominator of w and a.
110
Let r be (-4754)/574 + 6/2. Let l = r - 6/41. What is the common denominator of l and 1*1/6 - 2?
42
Let z = 3 + 3. Let l = 34 + -24. Calculate the smallest common multiple of l and z.
30
Let y = -2185 + 21811/10. Suppose -12 = -3*w, 5*w = 3*q - 12 - 10. Calculate the common denominator of -10*(-10)/q + -2 and y.
70
Let t = -297349639/22993652 + -2/522583. Let b = t + 288/11. What is the common denominator of b and -4/3?
12
Find the common denominator of 95/22 and 5/3*(-702)/(-144).
88
Let p = 46 + -22. Calculate the common denominator of (1 - 28/p)*-101 and -43/6.
6
Let j(t) = -3*t**3 + t**2. Let i be j(-1). Suppose 2*z + 47 = d, -5 - 1 = z - i*d. What is the common denominator of 2/8 + z/(-56) and 39/4?
28
Suppose 0 = 2*w - 3 - 19. Suppose 5*u - m - 20 = 0, -2*m - 2*m + 25 = u. Calculate the smallest common multiple of w and u.
55
Suppose 0 = 4*q - 3*y + 10107, -5*q - y - 12384 = 245. Let t be ((-34)/18)/(q/10845). Let f = t + 24/421. What is the common denominator of -40 and f?
6
Let r be (-1)/1 + 0 + 9. Find the common denominator of 4/3*(-189)/r and 27/8.
8
Let n(v) = v**3 + v**2. Let g(w) = -w + 5. Let l be g(3). Calculate the least common multiple of n(l) and 8.
24
Suppose 0 = -4*t + 3 + 1. Calculate the common denominator of (t - 4) + (-385)/84 and 43/10.
60
Let w = 5823/4 - 40611/28. What is the common denominator of 51/8 and w?
56
Let z(f) = f**3 - 4*f**2 - 5*f + 2. Suppose 4*n + 0*n - 5*t - 40 = 0, -17 = -n + 3*t. What is the smallest common multiple of 6 and z(n)?
6
Calculate the common denominator of 7/16 and (-8)/44 - (-5045)/1430.
208
Suppose -12*h + 182 = 38. Calculate the lowest common multiple of 44 and h.
132
Let r(f) = -f**2 - 16*f - 1. What is the lowest common multiple of 16 and r(-15)?
112
Let h = -3523 - -21077/6. What is the common denominator of h and 357/(-36) - 5/(-15)?
12
Let t(b) = -b**2 - 11*b - 9. What is the lowest common multiple of t(-10) and 2?
2
Let y = 55847/20 + -2795. What is the common denominator of 83/10 and y?
20
Let s = -3776 + 34024/9. Find the common denominator of s and ((-24)/(-3))/((-14)/11).
63
L


Hi,
Still struggling to get my Soekris to boot.
Here is the full boot sequence: http://paste.pocoo.org/show/509089/
Can't see anything obvious, is there any way to have logs more verbose?
Thanks
--
Jean-Marc


Effects of various saccharides on the masking of epitope sites and uptake in the gut of cedar allergen Cry j 1-saccharide conjugates by a naturally occurring Maillard reaction.
A major allergen of Japanese cedar, Cry j 1, was conjugated with galactomannan (M(w) of 15 kDa), dextran (M(w) of 12 kDa), xyloglucan (M(w) of 1.4 kDa), and various monosaccharides through the Maillard reaction by dry-heating in 65% relative humidity. The Cry j 1-galactomannan conjugate completely masked the epitopes of the allergen in Cry j 1. The Cry j 1-dextran conjugate also masked the epitopes of Cry j 1. The small size of oligosaccharide (xyloglucan) and various monosaccharides cannot mask the epitopes of allergen Cry j 1. This suggests that the higher molecular size of attached saccharides is important to mask sterically the epitope sites. The Cry j 1-galactomannan and Cry j 1-mannose conjugates were effectively trafficked in the gut and co-localized with immune cells, such as dendritic cells in the gut, suggesting that Cry j 1-saccharide conjugates are phagocytosed via the mannose receptor in immune cells. These results suggest that the Cry j 1-galactomannan conjugate is suitable for masking the epitope sites of Cry j 1 and trafficking to immune cells in gut lumen.


Chronic pulmonary artery balloon counterpulsation in sheep via percutaneous route.
In clinical practice pulmonary artery balloon counterpulsation (PABC) has been utilized only in the operative setting with the balloon housed in a graft attached to the pulmonary artery. Clearly, percutaneous insertion of a dedicated pulmonary artery balloon is a desirable goal for patients requiring temporary assist for right ventricular failure. To address the question of right sided cardiopulmonary tolerance for a chronic indwelling pulmonary artery balloon, six adult ewes underwent percutaneous placement of an 11 ml pulmonary artery balloon, via the femoral vein. Effective pumping and timing were monitored for 48 hours at which time the animals were sacrificed. At autopsy gross and microscopic study of all right heart structures, the pulmonary arteries and the lungs were studied for adverse effects. There were inconsequential minor abrasions to right heart structures in most animals. The pulmonary artery in five of six animals revealed ecchymoses and some transmural hemorrhage, but no necrosis or perforation. There was no pulmonary injury that could not be ascribed to postoperative atelectasis. This study demonstrates that chronic pulmonary artery balloon counterpulsation can be carried out for a period of 48 hours without significant injury to right heart and pulmonary structures in the ovine model


Phenotypic comparison of multiple monocyte-related populations in murine peripheral blood and bone marrow.
Monocyte subsets are not well defined in murine peripheral blood (PB). Monocyte-related populations could also be located in bone marrow (BM), but studies correlating monocyte populations found in these two tissues are lacking. This study simultaneously analyzed PB and BM to phenotypically define multiple monocyte-related subsets in each location. Murine PB and BM cells were simultaneously stained for monocyte-related populations, using five-color flow cytometry. Relevant subsets were defined on the basis of Ly-6C, CD11b, and wheat germ agglutinin phenotype in addition to light-scatter characteristics. These populations were extensively characterized for the expression of other myeloid and dendritic cell markers, adhesion molecules, chemokine receptors, and growth factor receptors. Six monocyte-related populations were defined, three each in BM and PB. No identical populations were found between the two tissues. Two BM populations and one PB population have heterogeneous expression of many markers, suggesting additional complexity among monocyte-related subsets. The murine monocytic series comprises multiple subsets, differing between PB and BM. This study defines and extensively phenotypes six of these populations, providing preliminary information about possible developmental relationships and migratory capacities of these cells.


Introduction
============

One of the salient features of eukaryotic genomes is the pervasive presence of introns, comprising up to 95% of transcribed primary protein-coding sequences in mammals \[[@B1]\]. The functions, origins and evolutionary trajectory of introns have long been of great interest in genomics and genome evolution. Although some theories of the spread of introns postulate that they can accumulate passively as a consequence of insufficient purifying selection to remove them in organisms with relatively low effective population sizes \[[@B2],[@B3]\], introns have been shown to play a number of functional roles \[[@B4]\]. They give rise to the possibility of alternative splicing, contributing to the diversity of biomolecules, and they can also have a substantial impact on levels of gene expression \[[@B5]-[@B7]\]. Introns contain most of the sequence features required for splicing (5' and 3' splice sites, branch point sites (BPS), poly-pyrimidine tracts and various intronic splicing elements). Many introns also contain functional non-coding RNAs \[[@B4]\], which can play critical roles in fine-tuning gene expression \[[@B8]\]. Lastly, introns have been proposed to control the timing of gene expression by delaying transcription \[[@B9]\].

Negative feedback loops with a time delay can result in oscillating patterns of gene expression, which can be exploited by living organisms as biological time-keeping devices. This appears to be particularly important in development \[[@B9]\]. The hairy and enhancer of split 7 (*Hes7*) gene is involved in the control of somite formation through oscillatory patterns of gene expression \[[@B10]\]. Recently, Takashima *et al.*\[[@B11]\] investigated *in vivo* the impact of removing the introns from the mouse *Hes7* gene. They found that expression of the mutant *Hes7* gene occurred approximately 19 minutes earlier than for the wild-type and that this reduction in gene expression delay resulted in abolition of oscillations and segmentation defects. Given that the delay associated with transcribing introns appears to play a crucial role in the functioning of this gene, the length of the introns may be evolving under purifying selective pressure. Moreover, further examples of genes with highly conserved intron content across species may reveal more genes for which transcriptional delay is an important aspect of gene regulation. We investigated the conservation of the intron content of *Hes7* across a diverse set of nineteen mammals and carried out a search for other genes for which total intron length shows evidence of evolutionary conservation. In such cases introns are likely to play an important functional role and, in a subset, time delays associated with transcribing introns may be a significant aspect of gene regulation.

Results and discussion
======================

Delayed expression of *Hes7* resulting, at least in part, from intron transcription has been proposed to play a key role in somite formation in animal embryos by establishing an oscillating pattern of gene expression \[[@B11]-[@B14]\]. Given the important role played by the length of the introns rather than their specific sequence content in this gene, we compared the combined length of *Hes7* introns across a diverse set of mammalian species (Fig. [1](#F1){ref-type="fig"}) to determine the extent to which the intron content of this gene is conserved over evolutionary time. For each orthologue, the sum of the distances between successive exons in the canonical transcript (according to Ensembl) was calculated. Despite the fact that there may be differences in gene annotation across species the intron content of most of the orthologues was similar. Nine out of the twelve of the available orthologues of *Hes7* among the mammalian species in our dataset differed in combined intron length from the human gene by less than 10% (Table [1](#T1){ref-type="table"}).

![**Phylogeny** Phylogenetic tree illustrating relationships of the 19 mammals included in this study.](1471-2105-12-S9-S16-1){#F1}

######

**Conservation of Hes7 intronlength** Intron content of orthologues of human *Hes7* (OrthoDB ID EOG45TDC4). The tarsier orthologue was omitted. This gene had no annotated introns but the annotation appeared to be incomplete.

  Species                      Common name                 Gene                 \#introns   CIL^1^
  ---------------------------- --------------------------- -------------------- ----------- --------
  *Homo sapiens*               Human                       ENSG00000179111      3           1839
  *Mus musculus*               Mouse                       ENSMUSG00000023781   3           1846
  *Oryctolagus cuniculus*      Rabbit                      ENSOCUG00000002606   3           1696
  *Canis familiaris*           Dog                         ENSCAFG00000016957   3           1823
  *Felis catus*                Cat                         ENSFCAG00000015190   3           1958
  *Equus caballus*             Horse                       ENSECAG00000013278   3           1804
  *Myotis lucifugus*           Little brown bat            ENSMLUG00000008014   3           1810
  *Erinaceus europaeus*        Hedgehog                    ENSEEUG00000006336   3           2774
  *Sus scrofa*                 Pig                         ENSSSCG00000017982   4           1550
  *Tursiops truncatus*         Bottlenose dolphin          ENSTTRG00000010312   3           1827
  *Loxodonta africana*         African elephant            ENSLAFG00000002331   3           1870
  *Monodelphis domestica*      Grey short-tailed opossum   ENSMODG00000007704   3           2364
  *Ornithorhynchus anatinus*   Platypus                    ENSOANG00000022456   2           2018
  *Bos taurus*                 Cow                         ENSBTAG00000012436   3           2364
  *Gorilla gorilla*            Gorilla                     ENSGGOG00000005760   3           1840

^1^Cumulative intron **length**

To determine whether intron content, and thus potentially transcription time, of *Hes7* was exceptionally conserved compared to other genes and to discover additional examples of genes for which there is evidence of intron content conservation we compared intron content (defined as the sum of the lengths of all introns in the canonical transcript) between human gene models from Ensembl and sets of orthologous genes, obtained from OrthoDB \[[@B15]\]. Because the pattern of insertion and deletion may differ between very large and smaller introns we restricted to genes with introns of similar lengths to *Hes7* (specifically we considered genes for which the sum of the intron lengths in the canonical transcript was between one and five kilobases). Restricting also to orthologous groups represented in at least half of the species, only 11 other orthologous groups (from a total of 1875 satisfying these criteria) had intron content within 10% of the human gene in as high a proportion of the orthologues as the *Hes7* group. Remarkably, of the corresponding 12 human genes (including *Hes7*), five are annotated with the gene ontology (GO) term GO:0007389 (pattern specification process) from the biological process component of GO, defined as "any developmental process that results in the creation of defined areas or spaces within an organism to which cells respond and eventually are instructed to differentiate". Using the DAVID functional annotation tool \[[@B16],[@B17]\], this is the most statistically enriched GO term (p = 9.4 × 10^--5^) and remains significant following correction for multiplicity of testing using the Benjamini-Hochberg method (FDR = 0.03). Six of the genes were annotated with the Swissprot and Protein Information Resource (SP-PIR) keywords term 'developmental protein' (FDR = 0.003).

We also used a more relaxed conservation threshold and identified a larger set of 144 genes with relatively conserved intron content among the genes with between one and five kilobases of intron. Genes for which more than 50% of the orthologues have intron content within 10% of the intron content of the human member of the group were included in this group, again restricting to orthologue groups represented in at least half of the species. We again found evidence for a set of genes with conserved intron content which was very highly enriched for genes involved in development, identified using DAVID (Table [2](#T2){ref-type="table"}). Only the top 20 enriched terms are shown in Table [2](#T2){ref-type="table"}. These include terms corresponding to the presence of homeobox DNA-binding domains, frequently involved in developmental gene regulation. To look for examples of conserved intron content among genes with higher intron content, we separated the human genes into four intron content size classes (1 -- 5 kb, 5 -- 20 kb, 20 -- 50 kb and 50 -- 100 kb), to account, to some extent, for the fact that patterns of intron length evolution may differ between genes with larger versus smaller introns. The proportion of genes from the other size classes (other than size class one, discussed above) that showed evidence of conserved intron content was smaller and we found no evidence for enrichment for the default DAVID gene sets (after multiplicity correction).

######

**Gene set enrichment analysis** Gene set enrichment analysis of genes in size class 1 showing evidence of intron content conservation in mammals. Top twenty most significantly enriched terms are shown.

  Category          Term                                                                   Count    ***P*****-value**   FDR (BH)^1^
  ----------------- ---------------------------------------------------------------------- -------- ------------------- -------------
  UP_SEQ_FEATURE    DNA-binding region:Homeobox                                            **20**   2 × 10^--10^        8 x 10^--8^
  INTERPRO          Homeobox, conserved site                                               **20**   9 × 10^--10^        2 x 10^--7^
  SP_PIR_KEYWORDS   Homeobox                                                               **20**   2 × 10^--9^         4 x 10^--7^
  INTERPRO          Homeobox                                                               **19**   5 x 10^--9^         6 x 10^--7^
  GOTERM_BP_FAT     Regulation of transcription, DNA-dependent                             **41**   7 x 10^--9^         1 x 10^--5^
  INTERPRO          Homeodomain-related                                                    **19**   1 x 10^--8^         8 x 10^--7^
  GOTERM_BP_FAT     Regulation of RNA metabolic process                                    **41**   1 x 10^--8^         8 x 10^--6^
  GOTERM_BP_FAT     Positive regulation of transcription from RNA polymerase II promoter   **19**   1 x 10^--8^         6 x 10^--6^
  GOTERM_BP_FAT     Positive regulation of biosynthetic process                            **27**   3 x 10^--8^         1 x 10^--5^
  GOTERM_BP_FAT     Positive regulation of cellular biosynthetic process                   **27**   3 x 10^--8^         1 x 10^--5^
  GOTERM_MF_FAT     Transcription regulator activity                                       **38**   7 x 10^--8^         2 x 10^--5^
  GOTERM_BP_FAT     Positive regulation of gene expression                                 **24**   8 x 10^--8^         2 x 10^--5^
  GOTERM_BP_FAT     Positive regulation of transcription, DNA-dependent                    **21**   8 x 10^--8^         2 x 10^--5^
  GOTERM_BP_FAT     Positive regulation of RNA metabolic process                           **21**   8 x 10^--8^         2 x 10^--5^
  GOTERM_BP_FAT     Positive regulation of macromolecule biosynthetic process              **25**   9 x 10^--8^         2 x 10^--5^
  GOTERM_BP_FAT     Positive regulation of macromolecule metabolic process                 **27**   1 x 10^--7^         2 x 10^--5^
  GOTERM_BP_FAT     Positive regulation of nitrogen compound metabolic process             **25**   1 x 10^--7^         2 x 10^--5^
  GOTERM_MF_FAT     Sequence-specific DNA binding                                          **25**   1 x 10^--7^         2 x 10^--5^
  SMART             HOX                                                                    **19**   2 x 10^--7^         8 x 10^--6^
  GOTERM_BP_FAT     Positive regulation of transcription                                   **23**   2 x 10^--7^         3 x 10^--5^

^1^Benjamini-Hochberg FDR corrected *P*-values

Intronic insertions and deletions along lineages leading to human and mouse
---------------------------------------------------------------------------

To investigate evolution of intron length through insertion and deletion in greater detail we considered changes in intron length along the human and mouse lineages following divergence from their last common ancestor. Genomic sequences at ancestral nodes of the mammalian phylogeny were obtained from Ensembl. These sequences were inferred using a probabilistic method that has been shown to have high accuracy for the inference of insertion and deletion events. We mapped insertion and deletion events to introns.

Conservation of intron content could result from the absence of insertion and deletion events or from balance between insertion and deletion. For the 11 genes that showed as much conservation as *Hes7* across the mammalian panel, we found qualitative support for both effects. The total amount of insertion and deletion was lower in these genes but there were also some genes with substantial insertion and deletions in balance (Fig. [2](#F2){ref-type="fig"}). This was particularly evident in the case of indels along the lineage leading to mouse, however, as these genes were selected on the basis of their conservation in a panel that included humans and mouse this result should be considered to be illustrative only.

![**Comparison of insertions and deletions** The sum of the insertions and deletions that have taken place in genes in size class 1 along the human (a) and mouse (b) lineages since divergence from their common ancestor. Blue points, corresponding to the genes that showed as much or more intron content conservation than *Hes7* are shown against a background of grey points, corresponding to all genes in the size class. All axes are truncated, but the truncated axes include all of the blue points. Grey points lying with cumulative insertions or deletions greater than **200** bp for human or **2000** bp for mouse are not shown.](1471-2105-12-S9-S16-2){#F2}

We also identified the complete set of genes in size class one that were annotated with a homeobox domain by Interpro \[[@B18]\]. To investigate whether the change in intron content since the common ancestor of human and mouse was less than expected, given the inferred numbers of indels we used a randomization procedure (*described in* Data and methods). We applied this procedure to the set of homeobox domain proteins and found that the change in intron content of these genes along both the human and mouse lineages was far less than expected, given the number of indels inferred to have occurred in each lineage and under the null assumption that the ratio of insertions and deletions and size distribution of these events did not differ for these genes compared to the rest of genes in the size class. In the case of human, the mean absolute change in intron content was 450 bp per gene, whereas the median of this value in the randomized data was 846 bp per gene. The randomized values were less than or equal to the observed value in 87 of 10,000 randomizations (one-tailed p = 0.009). In mouse the mean absolute change in intron length was just 9 bp per gene. The median of this quantity across randomizations was 398 bp per gene. The values in the random data were lower than or equal to the observed data in 33 of 10,000 randomizations (one-tailed p = 0.003).

Contribution of conserved sequence elements to intron length conservation
-------------------------------------------------------------------------

From our results it appears that there is selection to prevent large changes in intron content in specific sets of genes, notably in genes involved in development and especially in early embryonic patterning. This selection could result from the presence of *cis* or *trans* (e.g. non-coding RNAs) regulatory elements within introns of these genes, which would be disrupted by insertions and even more so by deletions. However, the presence of regulatory elements would tend to increase the intron length \[[@B19]\], yet we found good evidence for classes of genes with well conserved intron content only among the genes with lower intron content. To test the contribution of conserved functional elements within intronic sequences on the conservation of intron content we obtained conservaton scores, calculated using phastCons, from the UCSC genome database. phastCons scores for the intronic regions of *Hes7* are shown as figure [3](#F3){ref-type="fig"}. While there is evidence of functional elements, these are relatively few and short and seem unlikely to prevent changes in intron length over evolutionary time. More quantitatively, we compared phastCons scores (averaged across all intronic positions) between the 144 genes with evidence of conserved intron content and the remainder of the genes in the smallest intron size class. The difference in mean phastCons scores was not significant (p = 0.33, Wilcoxon rank sum test). The difference in the proportion of conserved sites (phastCons scores \> 0.95) was also non-significant (p = 0.69), suggesting that a greater proportion of conserved intronic functional elements is not the cause of the intron length conservation in these genes.

![phastCons scores across introns of the Hes7 gene](1471-2105-12-S9-S16-3){#F3}

To explain our observations, conservation acting on functional elements would, perhaps, have to be balanced with selection for rapid induction, as rapidly induced genes have been shown to have short introns \[[@B20]\]. If the conservation of intron content is a consequence of balance between conservation of intronic functional elements and selection either for efficiency \[[@B21]\] or for rapid induction, this balance appears to be particularly significant for development-associated genes. For these genes, expression precision at the critical developmental junctures in which patterns are established in the developing embryo may be particularly crucial. An intriguing role for introns that has recently gained experimental support is in the establishment of oscillating patterns of gene expression through the control of time delays in expression \[[@B11]\]. These authors attributed a 19 minute delay to the presence of introns. However, the introns of *Hes7* are relatively short (\< 2 kb) and RNA polymerase II processes nucleotides at a rate in the order of 2 kb per minute \[[@B22],[@B23]\]. This may suggest either the existence of other functional elements within the intron that caused the delay in transcription or specific properties of the intron that result in an exceptionally slow transcription rate.

Conclusion
==========

Some theories of intron evolution have focussed on the energy cost associated with introns. Reduced intron lengths in highly expressed genes were proposed to result from selection for efficiency \[[@B21]\]. In support of this model, greater selective efficiency associated with larger effective population sizes of unicellular versus multicellular organisms, is associated with shorter introns \[[@B2],[@B3]\]. Alternatively longer introns in highly and ubiquitously expressed genes may be associated with greater regulatory complexity and the preservation of regulatory elements in introns \[[@B19]\]. Selection to reduce energetic costs of transcription and for rapid induction of some genes as well as selection to preserve regulatory elements in highly regulated genes may all be important factors for intron evolution. However, comparison of genes with very conserved intron contents carried out here, suggests that there may be a substantial number of genes for which the size of the primary transcript is an important and conserved feature. These genes are enriched for key developmental processes, particularly the establishment of early embryonic patterns. Since time delays in gene induction have been shown to be important for some such genes, we propose that selection to conserve transcription time is an important factor in the evolution of intron lengths, particularly in development-associated genes. Our analysis involved a combination of a heuristic examination of intron content across a panel of mammalian species and a statistical randomization approach, which does not take into account all of the factors that may affect the fixation probabilities of insertions and deletions in introns. A better understanding of the evolution of intron sizes through insertion and deletion requires the development of evolutionary models of intron length evolution, though this is challenging because of the diversity of insertion and deletion events and the difficulty in modelling the constraints imposed by functional elements within the introns on the occurrence and size distribution of these events. If an appropriate model of neutral evolution by insertion and deletion can be derived, such a model could be used to identify and quantify purifying selection acting on intron lengths and, perhaps, to discover examples of positive selection acting on changes in intron length over a phylogeny or specific branches of a phylogenetic tree.

Data and methods
================

Complete sets of gene models were downloaded from Ensembl release 62 \[[@B24]\] via BioMart for 19 mammalian species. These were *Homo sapiens*, *Bos taurus*, *Canis familiaris*, *Ornithorhynchus anatinus*, *Equus caballus*, *Erinaceus europaeus*, *Gorilla gorilla*, *Loxodonta africana*, *Monodelphis domestica*, *Myotis lucifugus*, *Microcebus murinus*, *Mus musculus*, *Oryctolagus cuniculus*, *Sus scrofa*, *Spermophilus tridecemlineatus*, *Tarsius syrichta*, *Tursiops truncatus*, *Vicugna pacos* and *Felis catus.* Species were selected to sample a broad range of mammalian evolutionary history. A phylogenetic tree of these taxa, obtained from the interactive tree of life \[[@B25],[@B26]\] is provided for illustrative purposes (Fig. [1](#F1){ref-type="fig"}). For each Ensembl gene in each species we considered all annotated exons and calculated the total intronic content of the gene as the sum of the gaps between non-overlapping successive exons in the canonical transcript associated with the gene. Genes were divided into four classes, depending on the total intron content of the gene (1 -- 5 kb, 5 -- 20 kb, 20 -- 50 kb and 50 -- 100 kb). Orthologous groups of mammalian proteins were downloaded from OrthoDB \[[@B15]\]. We extracted the protein identifiers from OrthoDB corresponding to each of the mammalian species included in the study. Where more than one protein from a species was included in the same orthologous group, we selected one paralogue at random. For each orthologous group, with a representative in at least half of the mammalian species considered we determined the number of species in which the intron content, as defined above, was within 10% of the length of the intron content of the human gene. This was done separately for genes in different intron content classes. Functional analysis of genes with conserved intron content was carried out using DAVID \[[@B16],[@B17]\]. For each size class, the genes for which the intron content showed evidence of conservation was used as the foreground set and the complete set of genes in the size class was used as the background set.

Genomic multiple sequence ancestor alignments, inferred using the Ensembl Enredo-Pecan-Ortheus pipeline \[[@B27]\], were downloaded from the comparative genomics section of the Ensembl FTP site. Insertions and deletions were inferred for ancestral sequences included in these alignments using a branch transducer method that has been shown to achieve high accuracy \[[@B28]\]. This allowed insertions and deletions to be placed on branches of the mammalian phylogeny. For reconstructed insertion and deletion events we focussed on the branches leading from the common ancestor of the euarchontoglires (which includes rodents and primates) to humans and mouse. All insertion and deletion events occurring within introns (at least 20 bp from exon-intron) boundaries were identified, based on Ensembl gene models. In this case, intronic indels were defined as indels that occur within the boundaries of the gene but not within 20 bp of any annotated exon associated with the gene.

Randomization study of intron length evolution along the human lineage
----------------------------------------------------------------------

For each intron size class we used the complete set of insertion and deletion events within the introns to define the null expectation of events in that intron size class. To test for evidence of purifying selection acting on the intron content of a gene or a set of genes we considered all of the insertion and deletion events in the gene(s) under consideration and for each event sampled an insertion or deletion from the total set of events in the introns of genes in that class. Given a set of insertions and deletions randomly sampled in this way we calculated the change in intron content implied by this set of insertions and deletions (sum of insertion lengths minus the sum of the lengths of the deletions), separately along the human and mouse lineages. This was repeated 10,000 times and in each case the implied change in intron content in the randomized data was compared to the change in intron content, given the actual insertions and deletions inferred to have occurred. The number of times the absolute value of the change in intron content in the randomized data was less than or equal to the absolute value of the change in intron content in the observed data was used to calculate a p-value, with a separate p-value calculated for the human and mouse lineages. These p-values indicate the probability of observing as little or less variation in the intron content of a gene or set of genes, given the number of insertion and deletion events that have occurred and under the assumption that these indel events have the same distribution as events in other genes in the same intron size class.

Authors\' contributions
=======================

CS conceived the project. CS and PK performed the analysis and co-wrote the manuscript.

Competing interests
===================

The authors declare that they have no competing interests.

Acknowledgements
================

CS is funded by Science Foundation Ireland (grant number 07/SK/M1211b). PK is funded by the National University of Ireland Galway.

This article has been published as part of *BMC Bioinformatics* Volume 12 Supplement 9, 2011: Proceedings of the Ninth Annual Research in Computational Molecular Biology (RECOMB) Satellite Workshop on Comparative Genomics. The full contents of the supplement are available online at <http://www.biomedcentral.com/1471-2105/12?issue=S9>.


How to perform a re-scan of the blockchain

Performing a re-scan of the blockchain

This guide is a walktrough on how to perform a re-scan of the blockchain (this guide can be used for both XFUEL and XBY). This should be done when you've imported a private key and your balance doesn't match that of what the block explorer is showing. You can check your wallet address on the block explorer (XBY or XFUEL) to see if there is a mismatch in balances.


On this episode of the show Mark and Stuart run through some of the games they’ve been playing and cover a fair bit of recent news and controversies that have been swirling around the gaming interwebs. On this show we talk about Vampyr, God Of War, The Last Guardian, The Last Day of June, Dark Echo, Nex Machina and a few others.

You can find additional content by us over on the LGR website at www.lapsedgamer.com and you can get in touch with us via Twitter @lapsedgamer


vs.1.1

# DYNAMIC: "DOWATERFOG"				"0..1"
# DYNAMIC: "SKINNING"				"0..1"

#include "macros.vsh"

local( $worldPos, $worldNormal, $projPos );

alloc $worldPos
alloc $projPos

&SkinPosition( $worldPos );

;------------------------------------------------------------------------------
; Transform the position from world to view space
;------------------------------------------------------------------------------
dp4 $projPos.x, $worldPos, $cViewProj0
dp4 $projPos.y, $worldPos, $cViewProj1
dp4 $projPos.z, $worldPos, $cViewProj2
dp4 $projPos.w, $worldPos, $cViewProj3

;------------------------------------------------------------------------------
; Normal is based on vertex position
;------------------------------------------------------------------------------
&AllocateRegister( \$worldNormal );
&AllocateRegister( \$normalDotUp );

sub $worldNormal, $worldPos, $SHADER_SPECIFIC_CONST_6		; Normal = (Pos - Eye origin)
dp3 $normalDotUp, $worldNormal, $SHADER_SPECIFIC_CONST_7		; Normal -= 0.5f * (Normal dot Eye Up) * Eye Up
mul $normalDotUp, $normalDotUp, $cHalf
mad $worldNormal, -$normalDotUp, $SHADER_SPECIFIC_CONST_7, $worldNormal

&FreeRegister( \$normalDotUp );

; normalize the normal
&Normalize( $worldNormal );

mov oPos, $projPos

;------------------------------------------------------------------------------
; Fog
;------------------------------------------------------------------------------
&CalcFog( $worldPos, $projPos );

; base tex coords
mov oT1.xy, $vTexCoord0

; spotlight texcoords
dp4 oT0.x, $worldPos, $SHADER_SPECIFIC_CONST_1
dp4 oT0.y, $worldPos, $SHADER_SPECIFIC_CONST_2
dp4 oT0.z, $worldPos, $SHADER_SPECIFIC_CONST_3
dp4 oT0.w, $worldPos, $SHADER_SPECIFIC_CONST_4

local( $worldPosToLightVector, $distFactors );

alloc $worldPosToLightVector

sub $worldPosToLightVector, $SHADER_SPECIFIC_CONST_0.xyz, $worldPos

local( $distatten );
alloc $distatten
; $distatten = [ 1, 1/dist, 1/distsquared ]

; dist squared
dp3 $distatten.z, $worldPosToLightVector, $worldPosToLightVector

; oodist
rsq $distatten.y, $distatten.z

mov $distatten.x, $cOne

local( $dist );
alloc $dist
mul $dist.x, $distatten.z, $distatten.y

rcp $distatten.z, $distatten.z ; 1/distsquared

local( $endFalloffFactor );
alloc $endFalloffFactor

; ( dist - farZ )
sub $endFalloffFactor.x, $dist.x, $SHADER_SPECIFIC_CONST_5.w
; 1 / ( (0.6f * farZ) - farZ)
mul $endFalloffFactor, $endFalloffFactor.x, $SHADER_SPECIFIC_CONST_0.w
max $endFalloffFactor, $endFalloffFactor, $cZero
min $endFalloffFactor, $endFalloffFactor, $cOne

local( $vertAtten );
alloc $vertAtten
dp3 $vertAtten, $distatten, $SHADER_SPECIFIC_CONST_5
mul $vertAtten, $vertAtten, $endFalloffFactor

; Normalize L
&Normalize( $worldPosToLightVector );

; N.L
dp3 $worldNormal, $worldNormal, $worldPosToLightVector

; Modulate distance attenuation with N.L
mul oD0, $vertAtten, $worldNormal

; iris
dp4 oT3.x, $SHADER_SPECIFIC_CONST_8, $worldPos
dp4 oT3.y, $SHADER_SPECIFIC_CONST_9, $worldPos

free $dist
free $endFalloffFactor
free $worldPos
free $worldNormal
free $projPos
free $worldPosToLightVector
free $distatten
free $vertAtten


1. Field of the Invention
The present invention relates to sonar sound sources and more particularly to a surface deployed buoy that generates underwater sonic impulses by releasing acoustic charges.
2. Description of the Prior Art
Use of underwater explosive charges to provide sonic impulses is well known in the art. Echo location, sensor testing and signalling are some of the applications requiring underwater deployment of such sonic charges. In echo location the sonic charge is used in the vicinity of hydrophones to provide an impulse for echo location of submerged geologic structures or vessels. In sensor testing applications sonic charges are released at a fixed depth and distance to determine the performance characteristics of underwater sensors. Underwater charges can also be timed to explode in a coded pattern to communicate between a subsurface craft and a hydrophone.
Submarine launched acoustic charge release devices are often deployed for covertly signalling surface craft. The device used in covert signalling must be launched from a torpedo tube and provide coded release of signal charges. The device should not be visible from the surface. In one prior art device exploding bolts are used to separate the charge containing portion of the device from the control portion of the device. Charges are ejected in a coded sequence by explosive bolts in the charge containing portion of the device. A drogue is deployed from the control portion of the device to control the release of charges and prevent the ejected charge from damaging the control portion.
Aircraft deployed acoustic charge release devices are typically used in echo location applications. In this use, the explosion of a charge provides a sound source, and a sonobuoy dropped from the airplane act as pickup devices for echo location of submarines and underwater objects. Aircraft deployed devices are designed to release an underwater charge from a device containing a plurality of charges in response to a radio signal. The aircraft deployed device releases charges arrayed on a streamer to allow detonation when the buoy is signalled. The device has no provisions for recovery or reloading.
In test applications, charges are often deployed in the vicinity of an array of submarine acoustic sensors. Acoustic charges must be pressure sensitive to explode at a pre-set depth and the charges must be deployed within a standard distance from the array. The United States Navy has standard acoustic charges, known as signal underwater sound (SUS) charges, available for use in this application. If a surface vessel is used to deploy the charges, after deployment of a charge the vessel must be moved away from the charge deployment site to avoid interference with the acoustic signal because of echo from the hull or engine noise.
Each of the above methods of underwater acoustic charge deployment is deficient when deploying acoustic charges for test purposes. When testing acoustic sensors, charges must be released near the sensor with consistency. Neither aircraft nor submarines can release an acoustic source with sufficient accuracy for test purposes. Existing submarine and aircraft acoustic source devices do not allow for depth sensitive charges. Furthermore, aircraft and submarine deployment is too expensive to be practical for test purposes. Releasing acoustic charges from surface vessels is unavailable because of echo caused by the presence of the vessel and engine noise.
One prior art device correcting these deficiencies is Vaccaro et al., U.S. Pat. No. 5,175,712. The device of Vaccaro '712 provides for launch of acoustic charges at preselected time intervals from an autonomous craft but does not provide for acoustic charge launch on command from a platform nor any means of informing a recovery platform that all charges have been launched. Because of these deficiencies, the crew retrieving the '712 buoy does not know whether a charge remains in the buoy. Furthermore, the electrical door latch system of the '712 patent is unreliable and requires an excessive amount of power.


#include <swap.h>
#include <swapfs.h>
#include <mmu.h>
#include <fs.h>
#include <ide.h>
#include <pmm.h>
#include <assert.h>

void
swapfs_init(void) {
    static_assert((PGSIZE % SECTSIZE) == 0);
    if (!ide_device_valid(SWAP_DEV_NO)) {
        panic("swap fs isn't available.\n");
    }
    max_swap_offset = ide_device_size(SWAP_DEV_NO) / (PGSIZE / SECTSIZE);
}

int
swapfs_read(swap_entry_t entry, struct Page *page) {
    return ide_read_secs(SWAP_DEV_NO, swap_offset(entry) * PAGE_NSECT, page2kva(page), PAGE_NSECT);
}

int
swapfs_write(swap_entry_t entry, struct Page *page) {
    return ide_write_secs(SWAP_DEV_NO, swap_offset(entry) * PAGE_NSECT, page2kva(page), PAGE_NSECT);
}


Tricia Severns, ANP-BC, OCN®

Tricia (Sheehy) Severns has been an oncology nurse for more than 20 years. She graduated from the University of Pennsylvania School of Nursing in 1995. She returned to Boston in 1999 where she joined the Dana-Farber Inpatient Hospital at Brigham and Women’s Hospital. She joined the Dana-Farber Outpatient staff in 2001, starting in the infusion room. She returned to graduate school for a Masters’ in Nursing at MGH-Institute of Health Professions, and graduated with a 4.0 GPA. She served as a Waldenstrom’s Macroglobulinemia Nurse Practitioner from 2008-2103 under the direct supervision of Dr. Steven Treon. She had the opportunity to present at IWMF meetings, as well as WM Summits. She took a brief hiatus and rejoined Dana-Farber in 2015 where she became a Bone Marrow Biopsy Proceduralist. She was invited back to the WM team in February 2016, where she now works part-time in addition to her role in the Procedure Suite. She provides direct patient care to patients with Waldenstrom’s Macroglobulinemia and related diseases.


Gold stocks are set to boom thanks to the commodity's price and merger activity, with UBS slapping a buy on its entire gold coverage list with the exception of Newcrest and Independence Group.

UBS put out the note as the sector enjoys a boost from a wave of mergers and acquisitions this year, with 26 deals worth in total $1.7 billion.

Gold has recovered from its March lows of $US1150 an ounce and is now trading near $US1188. Credit:Eddie Jim

Gold has recovered from its March lows of $US1150 an ounce and is now trading near $US1188 ($1533).

"With the exception of Newcrest and Independence Group, we have buy ratings on our entire gold coverage list," said UBS.


GOP strategy: Come up with ideas not good enough to persuade the Democrats; anything else is good

That’s the lesson to learn from Mitt Romney, who Jonathan Chait notices was endorsed by the National Review until the Democrats decided he made good policy ideas. Once Obama, Reid, and Pelosi adopted Romney’s health care plan, it was considered anathema and pretty much killed Romney’s chances of winning the Republican primary in 2012. Maybe that’s why the Republicans seem to be bereft of ideas. If they somehow persuade the Democrats to adopt them, they will be dirty. Or something.


Are we there yet?
Statistical approaches to Artificial Intelligence are behind most success stories of the field in the past decade. The idea of generating non-trivial behaviour by analysing vast amounts of data has enabled recommendation systems, search engines, spam filters, optical character recognition, machine translation and speech recognition, among other things. As we celebrate the spectacular achievements of this line of research, we need to assess its full potential and its limitations. What are the next steps to take towards machine intelligence?


BILLIONAIRE wealth in the UAE surged to $132 billion this year from $45 billion in 2013, marking a worldwide record increase of 193 per cent, a report said.

The number of billionaires in the UAE also increased to 46 from 37 in 2013, added the Wealth-X and UBS Billionaire Census 2014 released by Wealth-X, the definitive source of intelligence on the ultra-wealthy.

Meanwhile, Saudi Arabia saw a decline of 18.6 per cent in its billionaire wealth, down to $166 billion in 2014, as against $204 billion last year, according to the report, which noted that the kingdom claimed the 10th spot globally for its count of 57 billionaires.

The Middle East experienced a decline in its billionaire population over the past year.

Africa is the only other region that saw a decrease in its number of billionaires.

Nonetheless, the remaining billionaires, as well as some of the region’s new billionaires, have seen their total wealth increase. Only 6 per cent of the Middle East’s billionaires are female, the smallest share of any other region.

Only 10 per cent of the region’s billionaires fully inherited their wealth. Work, entrepreneurialism and family businesses are the most signiﬁcant factors behind the growth of the region’s billionaire wealth, the report said.

Asia, however, boasted the largest billionaire wealth increase, with the region’s billionaires’ fortunes growing by 18.7 per cent over the past year. The region is responsible for 30 per cent of the net increase in global billionaire wealth in 2014. Asia’s billionaire population grew by 10 per cent in 2014, with 52 new entrants into the billionaire club – 33 are from China.

The US maintains its position as the world’s top billionaire country with a population of 571 billionaires in 2014, followed by China (190) and the UK (130), which took the third spot from Germany (123) on the Top 40 Billionaire Countries/Territories list.

Europe is home to more than a third of the world’s billionaire population.

The billionaire population in the Middle East shrank by 1.9 per cent, but total billionaire wealth in the region rose by 16.7 per cent.

There are 2,039 male billionaires in 2014, accounting for 87.7 per cent of the world’s total billionaire wealth of $7.3 trillion.

There are 286 female billionaires in 2014, accounting for a 12.3 per cent share of global billionaire wealth.

Gulf banks show solid growth

GULF banks have shown healthy earnings growth over the last year and a half despite historically low interest rates, according to a new Standard & Poor’s (S&P) Ratings Services report.

The declining credit losses will continue to support GCC banks’ earnings throughout 2014, although this effect is expected to be less visible than in the last year, it said.

Timucin Engin, S&P’s analyst, said: “Prospects for economic growth in the Gulf region remain healthy for the next few years. We expect most Gulf banks to continue to benefit from robust corporate activity and consumer consumption over the next 18 to 24 months. The many infrastructure projects planned in the Gulf should translate into sustained streams of corporate lending.”

Over the past three years, strong liquidity flows into the Gulf’s deposit markets have supported the region’s banks, which traditionally rely on local deposits for the bulk of their funding, said the report.

Petchem firms boost corporate earnings

THE combined net profits of Saudi Arabia’s leading companies are expected to rise 17 per cent this year and a further 11 per cent in 2015, largely on the back of petrochemical producers and banks, although a number of companies in other sectors also promise strong growth.

The figures are based on average forecasts from analysts surveyed by Reuters for 81 companies which accounted for 99 per cent of the total profits of constituent companies in Saudi Arabia’s main equities index last year.

The outlook puts Saudi Arabia roughly on a par with Qatar, where profit growth is set to average 13 per cent this year and next, and Abu Dhabi, which is expected to average 17 per cent.

It’s a big improvement from Saudi Arabia’s zero profit growth in 2012 and a six per cent increase last year. That poor performance was almost entirely due to weak profits at petrochemical firms, which saw product prices sag because of the economic slump in Europe.

Strong earnings will not necessarily translate into further fast rises for Saudi stock prices, which are already up 28 per cent year-to-date, having surged since the market opening plan was announced in July; many fund managers think the market is no longer cheaply valued.

But the earnings picture does ensure that the market opening is likely to attract heavy investor activity.

UAE residents ‘save outside the country’

NEARLY half of UAE residents choose to save their money in bank accounts outside their country of residence, a report said.

More than 48 per cent of respondents of an annual survey conducted by compareit4me.com, a UAE-based finance comparison website, said they are more confident with banking systems in their home countries.

Jon Richards, CEO of compareit4me.com, said the results indicate that UAE financial institutions may be missing out on an opportunity and need to do more to keep residents’ savings onshore.

“When you look at the results, a significant number or 65.5 per cent of the survey respondents claimed to be saving between 10 per cent and over 50 per cent of their salary every month. With millions of dirhams leaving the country every year, there appears to be huge potential for UAE banks to attract regular savers,” Richards added.

The banks need to work with the UAE Central Bank to look at initiatives that will incentivise expat workers to keep money earned here in the country. At present, many expats feel more comfortable saving with banks in their home country as they understand the system and feel comfortable with banking regulations. Accounts in the Isle of Man are protected by the Compensation of Depositor Regulations.

Despite the large number of residents sending money abroad, renewed economic confidence in recent years has seen UAE banks flush with liquidity. According to the UAE Central Bank, total bank deposits of UAE banks climbed by 1.4 per cent year-on-year in June to Dh1.4 trillion ($381 billion).

Gulf Industry, published by Al Hilal Group has now established itself as a "must-read" publication for anyone with an involvement in the region's industrial sector whether as manufacturers, distributors or related service organisations.The publication represents a "window" into the Gulf's manufacturing trading and export sectors providing news, views analysis and information across virtually every industry related sector.


                         UNPUBLISHED

UNITED STATES COURT OF APPEALS
                FOR THE FOURTH CIRCUIT

UNITED STATES OF AMERICA,              
                 Plaintiff-Appellee,
                 v.                            No. 02-4966
ANTHONY TERRELL GARNER,
             Defendant-Appellant.
                                       
           Appeal from the United States District Court
         for the District of South Carolina, at Orangeburg.
                 Cameron M. Currie, District Judge.
                             (CR-01-945)

                      Submitted: May 12, 2003

                       Decided: June 2, 2003

         Before NIEMEYER and LUTTIG, Circuit Judges,
              and HAMILTON, Senior Circuit Judge.

Affirmed by unpublished per curiam opinion.

                            COUNSEL

Allen B. Burnside, Assistant Federal Public Defender, Columbia,
South Carolina, for Appellant. J. Strom Thurmond, Jr., United States
Attorney, Stacey D. Haynes, Assistant United States Attorney,
Columbia, South Carolina, for Appellee.
2                      UNITED STATES v. GARNER
Unpublished opinions are not binding precedent in this circuit. See
Local Rule 36(c).

                              OPINION

PER CURIAM:
   Anthony Terrell Garner pled guilty to two counts of possession of
a firearm by a convicted felon, 18 U.S.C. § 922(g)(1) (2000) (Counts
1 and 2), and one count of possession with intent to distribute mari-
juana, 21 U.S.C. § 841(b)(1)(D) (2000) (Count 4). He received con-
current sentences of 120 months for the firearms offenses and sixty
months for the marijuana offense. Garner appeals his sentence, argu-
ing that the district court erred in making a four-level enhancement
pursuant to U.S. Sentencing Guidelines Manual § 2K2.1(b)(5) (2001),
after finding that he possessed a firearm in connection with another
felony offense. We affirm.
   On January 1, 2001, Garner was arrested on a number of state
charges, including unlawful possession of two firearms, one of them
a stolen pistol. On April 3, 2002, a federal search warrant was exe-
cuted at the mobile home where Garner lived with his girlfriend. Law
enforcement officers seized 31.4 grams of marijuana packaged for
sale and stored in a zip-lock plastic bag, another 54.95 grams of mari-
juana packaged for sale and stored in a Crown Royal bag, and a small
amount of cocaine that Garner had in his pants pocket. They also
found an unloaded .380 caliber pistol under the mattress of Garner’s
bed. Garner admitted that the gun and the marijuana were his and pled
guilty to the instant offenses.
   At sentencing, the district court determined that Garner possessed
the gun seized from his residence on April 3 in connection with his
marijuana dealing, despite Garner’s testimony that he bought the gun
for the sole purpose of selling it and did not sell drugs from his home.
A factor in the court’s decision was Garner’s statement at the sentenc-
ing hearing that he had in the past carried a gun for protection while
selling marijuana and later sold that gun.
   Because Garner admitted that the gun seized on April 3 belonged
to him, the only issue on appeal is whether the district court correctly
                       UNITED STATES v. GARNER                          3
determined that he possessed it in connection with his marijuana
offense. The parties agree that the phrase "in connection with," as
used in § 2K2.1(b)(5), has been held by this Court to have the same
meaning as it does in § 2K2.1(c)(1), and that both phrases are analo-
gous to the "in relation to" language in 18 U.S.C. § 924(c) (2000). See
United States v. Garnett, 243 F.3d 824, 828 & n.6 (4th Cir. 2001).
The district court’s factual findings with respect to applicability of the
enhancement are reviewed for clear error. Id. at 828. To prove that a
firearm was possessed "in relation to" another felony, the government
must show that the gun had "some purpose or effect with respect to"
the other felony, and facilitated or had the potential to facilitate the
other offense. United States v. Lipford, 203 F.3d 259, 266 (4th Cir.
2000) (quoting Smith v. United States, 508 U.S. 223, 237 (1993)). The
presence of the firearm may not be simply accidental or coincidental,
but it is sufficient "if the firearm was present for protection or to
embolden the actor." Id. (citing United States v. Mitchell, 104 F.3d
649, 654 (4th Cir. 1997)).

   Garner argues that no reliable evidence established a connection
between his drug sales and the pistol found in his home. He suggests
that his case is like United States v. Wilson, 115 F.3d 1185, 1191-92
(4th Cir. 1997) (reversing § 924(c) conviction where informant who
approached the defendant to buy marijuana bought rifle instead,
because there was no evidence that the rifle facilitated or had the
potential to facilitate a drug offense). Wilson is distinguishable from
Garner’s case, however, because Garner admitted that he had used a
gun to protect himself while selling drugs. The gun seized from Gar-
ner’s house had the potential to facilitate his drug dealing. Garner
could easily have obtained ammunition for the gun and, even
unloaded, the gun could have been used to intimidate. Therefore, the
district court did not clearly err in finding that Garner possessed the
gun in connection with his drug business.

   We therefore affirm the sentence imposed by the district court. We
dispense with oral argument because the facts and legal contentions
are adequately presented in the materials before the court and argu-
ment would not aid the decisional process.

                                                             AFFIRMED


1943 St. Louis Browns season

The 1943 St. Louis Browns season involved the Browns finishing 6th in the American League with a record of 72 wins and 80 losses.

Offseason
 November 7, 1942: Hal Epps was drafted by the Browns from the St. Louis Cardinals in the 1942 minor league draft.
 February 1, 1943: Paul Dean, brother of Browns broadcaster Dizzy Dean, was purchased by the Browns from the Washington Senators.

Regular season

Season standings

Record vs. opponents

Roster

Player stats

Batting

Starters by position
Note: Pos = Position; G = Games played; AB = At bats; H = Hits; Avg. = Batting average; HR = Home runs; RBI = Runs batted in

Other batters
Note: G = Games played; AB = At bats; H = Hits; Avg. = Batting average; HR = Home runs; RBI = Runs batted in

Pitching

Starting pitchers
Note: G = Games pitched; IP = Innings pitched; W = Wins; L = Losses; ERA = Earned run average; SO = Strikeouts

Other pitchers
Note: G = Games pitched; IP = Innings pitched; W = Wins; L = Losses; ERA = Earned run average; SO = Strikeouts

Relief pitchers
Note: G = Games pitched; W = Wins; L = Losses; SV = Saves; ERA = Earned run average; SO = Strikeouts

Farm system

Awards and honors
Vern Stephens, American League All-Star

References

External links
1943 St. Louis Browns team page at Baseball Reference
1943 St. Louis Browns season at baseball-almanac.com

Category:St. Louis Browns seasons
Saint Louis Browns season
St Louis Browns


The Big 3

August 16, 2018

Have you ever felt that some things are just too impossible to change? That they’ll never change, that they’ll never heal or that relationship will never be the same? We all have these impossible mountains around us that we’ve learned to accept as immovable.

But then again, what if they’re not?

The gospels record three miracles of Jesus literally raising someone from the dead… What if he can still do the impossible? Join us as we discuss The Big 3 miracles of life change in the Bible and explore how God still brings dead things back to life.


Receive the latest local updates in your inbox

Police say a 69-year-old man has been taken into custody after he allegedly stabbed his 59-year-old girlfriend to death during an argument in their Greenwich Village apartment Thursday. Checkey Beckford reports. (Published Friday, May 2, 2014)

A 69-year-old man is accused of killing his 59-year-old girlfriend with an ax and a knife during an argument in their Greenwich Village apartment Thursday, police say.

Dennis Guglielmo has been charged with second-degree murder and criminal possession of a weapon after the death of Alice Birnbaum.

Police say the suspect and victim were arguing in their third-floor apartment on Downing Street when the man allegedly attacked her with both an ax and knife, police say.

Birnbaum had multiple stab wounds to her chest and body, according to police. She was pronounced dead at the scene, and Guglielmo was taken into custody.

Guglielmo is a former co-owner of Carmine Street Guitars, a music store where musicians pay $1,500 for custom guitars made from wood salvaged from landmarks around the city, the New York Times reports. He lost his stake in the store after unsuccessfully suing his partner, a fellow business owner told the Times.

Stunned neighbors said the suspect is well-known and liked in the area

Attorney information for Guglielmo wasn't immediately available.

-- Checkey Beckford contributed to this report.

Published at 9:29 PM EDT on May 1, 2014 | Updated at 7:42 AM EDT on May 2, 2014


Dallas Cowboys can clinch NFC East with win over Giants

The Cowboys’ Dak Prescott (4), Ezekiel Elliott (21) run on to the field before a recent game.

The Cowboys’ Dak Prescott (4), Ezekiel Elliott (21) run on to the field before a recent game.

Photo: The Associated Press File

Photo: The Associated Press File

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The Cowboys’ Dak Prescott (4), Ezekiel Elliott (21) run on to the field before a recent game.

The Cowboys’ Dak Prescott (4), Ezekiel Elliott (21) run on to the field before a recent game.

Photo: The Associated Press File

Dallas Cowboys can clinch NFC East with win over Giants

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EAST RUTHERFORD, N.J. >> The Dallas Cowboys have everything in front of them, and one little nagging thing behind them: the New York Giants.

The winners of 11 straight games, the Cowboys (11-1) can clinch the NFC East and move a step closer to nailing down home-field advantage for the conference playoffs if they can beat the Giants (8-4) at MetLife Stadium on Sunday night.

While the Cowboys insist this is just another game, you know there has to be some churning inside them in facing their division rivals.

After all, the Giants are the reason Dak Prescott, Ezekiel Elliott and company aren’t perfect.

New York posted a 20-19 win in Dallas in the season opener, and one has to believe Jason Garrett’s crew wants payback. And if not that, the Cowboys certainly don’t want to go into the playoffs wondering if they can beat the Giants.

“I’m looking forward to it a lot personally, and then I know this team,” Prescott said. “They’re the only blemish on our record right now, and just to be able to go up there at their place and be able to do what they did to us the first game, we’re excited for the opportunity.”

The Giants know the Cowboys will be motivated.

“They’re on a roll right now and I’m sure they don’t want to lose,” Giants receiver Odell Beckham Jr. said. “We’re the only team that has beaten them. I’m sure that that’s on their mind, too. They feel that they could be 12-0. They’re going to try and come out and prove that they should be undefeated. It’s going to be a big challenge.”

The challenge got bigger for the Giants on Wednesday when defensive end Jason Pierre-Paul was lost for at least the rest of the regular season after having surgery to repair a sports hernia.

Not only did the team lose one of its playmakers on defense, it came in a week when the Giants are facing new questions after a poor performance in Pittsburgh ended their six-game winning streak. They still hold the top spot in the wild-card playoff chase, but their schedule is tough with Detroit, Philadelphia and Washington in the final three weeks.

“We believe we are a really good team, even though we haven’t played our best ball yet,” Giants linebacker Jonathan Casillas said. “If we can put the game together this week, I think it will be a statement game, but we have to do it. We have to be able to win, and that’s easier said than done.”

Here are some things to watch in the nationally televised game:

REPLACING JPP: Free-agent rookie Romeo Okwara probably will start at left defensive end, with Kerry Wynn and Owa Odighizuwa backing him up. Don’t be surprised if the Giants occasionally flip Olivier Vernon to the left side just to give the Cowboys something to think about. Linebacker Devon Kennard might see time as a stand-up speed rusher. Wynn is the best run stopper of the group, so he may need to get more time if the Giants want to limit Elliott, who leads the league in rushing.

RESPECT FROM ODB: Beckham knows a thing or two about starring right away after winning the 2014 Offensive Rookie of the Year award. So he’s highly complimentary of Prescott and Elliott. “Absolutely, on the Giants or not on the Giants, you have to respect that,” Beckham said.

Beckham, a former LSU star, remembers facing Mississippi State’s Prescott when they were in the SEC, and he offered Elliott advice on life in the NFL before the season. “I love to see Zeke out there celebrating,” Beckham said. “I love to see Dak out there with a smile. You just want to minimize that this Sunday.”

GIANTS OFFENSE: The offense has been inconsistent. It has yet to score 30 points. The running game is among the worst in the league and Beckham has been the only playmaker for Eli Manning. Against the Steelers, Victor Cruz was not targeted once. The week before, Sterling Shepard wasn’t targeted against Cleveland.

“We need to score more points and be more productive. Just have to overall execute better,” Beckham said.

LEADING 88?: Dez Bryant’s next receiving touchdown will push him past Michael Irvin for second on Dallas’ career list. Both have 65. Irvin is the “Hall of Famer 88,” who followed the “Original 88” in Drew Pearson, who had 48 touchdowns. The franchise leader is another Hall of Famer, Bob Hayes, with 71.

“He gets that confidence going and I have all the confidence in him,” Prescott said. “He feels no one can guard him and I feel the same way.”

STOPPING ELLIOTT: The Giants held Elliott to a season-low 51 yards on 20 carries in his first game. The key is in the middle with tackles Damon Harrison and Johnathan Hankins. “From the very first time we played against him, we all said that that guy is going to be a player in this league, and that seems to be true up until this point,” Harrison said.


Infant race affects application of clinical guidelines when screening for drugs of abuse in newborns.
Screening for illicit drugs in newborns has privacy, social, and legal risks for families of the infants. Established drug-screening criteria may be applied in a manner that considers nonproven risk factors such as race in addition to evidence-based factors. The goal of this study was to determine if race was used as a criterion for screening infants for intrauterine cocaine exposure. We hypothesized that infants of black mothers would be more likely to be screened regardless of whether they met the standard criteria for screening of our institution's NICU. We used the electronic medical records of newborn infants and their mothers to determine which mother-infant pairs had documented evidence of meeting the criteria for screening infants for prenatal exposure to illicit drugs set forth in the guidelines of our NICU. We then assessed the rates of drug screening to determine the strongest predictors of whether an infant would be screened. We assessed 2121 mother-infant pairs. Infants born to black mothers were more likely than those born to white mothers to have screening performed whether they met screening criteria (35.1% vs 12.9%; P < .001) or did not (5.3% vs 1.2%; P < .001). In a logistic regression analysis, black race remained independently associated (odds ratio: 2.17 [95% confidence interval: 1.25-3.79]) with drug screening even when we controlled for our standard screening criteria and income, insurance status, and maternal education. Providers seemed to have used race, in addition to recognized risk criteria, as a factor in deciding whether to screen an infant for maternal illicit drug use.


The K-2 submarine, designed as the Soviet Union’s next generation heavy submarine, would have combined many of these technologies into a weapon worthy of a James Bond villain. From V-2 rockets to tank-carrying midget submarines, the super sub could have done it all. Although the K-2 class was never built, it remains a window into the triumphant visions of the Soviet Union and Stalin itself in the aftermath of the most destructive war in history.

In the aftermath of World War II, both the U.S. and the Soviet Union raced to recover Germany technology in a variety of military fields, including submarines and rocketry. The American effort, known as Project Paperclip, followed the Allied forces on the Western Front as they swept into France, the Low Countries, and eventually Germany itself, capturing technical documents, working equipment, and even the scientists and engineers that worked on these secret projects.

In an article on his Covert Shores website, submarine expert HI Sutton explains the Soviet effort to use many of these captured technologies to build a super submarine.

P-2 submarine. HI Sutton/Covert Shores

In 1949, Soviet designers proposed a submarine that incorporated a number of these technologies. The sub, known as K-2, would be 367 feet long, 41 feet wide, and displace approximately 5,360 tons, making it easily ten times larger than wartime submarines constructed just four years before. K-2 had a crew of 100, a maximum operating depth of 656 feet, and a maximum speed of 17 knots underwater.

The K-2 itself, Sutton explained to Popular Mechanics, was heavily influenced by the revolutionary German Type-XXI U-boat of World War II. The Type XXI was the most advanced submarine fielded by any country during the war, and details of the submarine had fallen into Soviet hands.

K-2 was also armed to the teeth, almost comically so. “The submarine was massive for the time,” Sutton says, “with sixteen torpedo tubes in three separate torpedo rooms (two forward, one aft), all with reloads. This compared to ten on U.S. Navy fleet submarines of this time, and just four on later U.S. nuclear-powered attack submarines.”

Each fired ET-46 torpedoes. It also featured two pairs of anti-aircraft guns, both 57-millimeter and 25-millimeter, for fending off sub-hunting aircraft. All of this was pretty standard for submarines of the late 1940s.

German V-2 rocket preparing for launch. Fox Photos Getty Images

Amazingly, the torpedoes and guns were actually K-2’s secondary armament. Primary armament consisted of no less than 12 R-1 (NATO code name: SS-1 “SCUNNER”) ballistic rockets, copies of the German V-2 rocket , in a large payload bay. Alternately, the payload bay could accommodate up to 41 10X (“SWALLOW”) cruise missiles, copies of the German V-1 cruise missile . Alternately, for a completely different role it could nine midget submarines, each carrying tanks for amphibious invasions.

“P2 was the height of ambition of Stalin's navy,” Sutton explained. “It combined the latest submarine design learnings from World War II with the latest missiles. Until the Soviets captured Nazi German technologies at the end of World War Two, they did not have access to such advanced missile concepts.”

V-2 buzz bomb, known as SWALLOW in Soviet service. Keystone-France Getty Images

Why would Russia build a submarine suitable for wildly different goals as bombarding countries with rockets and cruise missiles versus landing amphibious forces?

“The ability to depot tank-carrying midget submarines shows the confused state of Soviet requirements," says Sutton. "Transport submarines and use of submarines firing amphibious operations was seriously considered on both sides of the iron curtain, but mixing amphibious transport and missile strike capabilities in a single submarine is difficult to explain.”

One possible explanation for the do it all super sub: Stalin himself. The Soviet leader, like Adolf Hitler, had a tendency to meddle in military decision making, and his megalomania could have persuaded him to pursue such an idea.

Alternately, the Soviet Navy could have pitched the submarine as a wonder weapon to help ensure it was funded. The more useful a weapon is in a variety of roles the more likely it will stand out as worthy of state funding.

The German Type XXI submarine, which formed the technical basis for the K-2. DEA PICTURE LIBRARY Getty Images

But for whatever the reason, the K-2 was never built. “Russia later built separate ballistic missile submarines and cruise missile submarines," Sutton told PM. "No amphibious transport submarines were ever built.”

Soviet progress in missile submarines wisely went a more incremental route, leading to monsters such as the Typhoon and Oscar classes in the 1980s—submarines that were several times bigger than K-2.

Source: Covert Shores

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