# Totalling Troublesome T's

Given a "T" shape on an x * y number grid, with length W on the top bar and H on the stem of the T, with the bottom of the T on the square numbered n: calculate the total of all of the numbers in the T shape. W must be an odd number and all must be positive integers.

Here are some examples of valid T's on a 9*9 grid:

Looking at the T where n=32, W=3 & H=4, you can see that the total is:
4 + 5 + 6 + 14 + 23 + 32 = 84.

# The Challenge

Your challenge is to create a program which, when given five positive integers, x, y, W, H and n, output the total of the T with those values (W must be odd, given an even number the program can output anything or even crash). The numbers may be inputted in any reasonable format. If the T does not fit on the given grid, then any output is acceptable.

# Example Code (Python 3.6.2)

x = int(input())
y = int(input())
W = int(input())
H = int(input())
n = int(input())

total = 0

#"Stem" of the T
for i in range(H - 1):
total += n - (x * i) #Decrease increment by length of row each time

#Top of the T
for i in range(-W//2 + 1, W//2 + 1): #Iterate over the width offset so that the center is zero
total += (n - (x * (H - 1))) + i #Add the top squares on the T, one less to the left and one more to the right

print(total)


As with most challenges here this is , so the answer with the shortest code in bytes wins.

• Isn't y redundant information? – Jonathan Allan Sep 6 '17 at 20:09
• @JonathanAllan I'm presuming it needn't be a square grid in which case y would not be redundant. x & y together would define the grid, n defines the bottom of the T shape, and W & H together define the T shape's size. – Engineer Toast Sep 6 '17 at 20:11
• @EngineerToast we are told n and x and told "If the T does not fit on the given grid, then any output is acceptable." so y may be inferred. – Jonathan Allan Sep 6 '17 at 20:12
• There could be many ways to approach the challenge, there's probably a builtin for it in Mathematical. – Leo Sep 6 '17 at 20:33
• Can you add some test cases, please? – Shaggy Sep 6 '17 at 22:19

# Jelly, 10 bytes

A construction technique, which ends up much like Luis Mendo's MATL answer

Ḷ×ạ⁵µṪ×⁶+S


A full program taking H, x, n, W in that order (y may be appended if one so wishes).

Try it online!

### How?

Ḷ×ạ⁵µṪ×⁶+S - Main link: H, x
Ḷ          - lowered range                         [0,1,2,...,H-2,H-1]
×         - multiply (vectorises)                 [0,x,2x,...,(H-2)x,(H-1)x]
⁵       - program's 5th argument (3rd input)    n
ạ        - absolute difference (vectorises)      [n,n-x,n-2x,...,n-(H-2)x,n-(H-1)x]
-  (note: for an in-range T n>(H-1)x)   (cells of the stem)
µ      - monadic chain separation (call this stem)
Ṫ     - tail (pop from AND modify stem)       n-(H-1)x
-                                       (cell at intersection of stem and top)
⁶   - program's 6th argument (4th input)    W
×    - multiply                              Wn-W(H-1)x
-                                       (total value of the top)
S - sum (the modified stem)               n+n-x+n-2x+...+n-(H-2)x
-                                       (total value of the stem w/o top cell)


## JavaScript (ES6), 32 30 bytes

Saved 2 bytes thanks to @Shaggy

(w,h,x)=>Q=n=>--h?n+Q(n-x):n*w


Takes input in a curried format: f(W,H,x,y)(n)

let f =
(w,h,x)=>Q=n=>--h?n+Q(n-x):n*w;

console.log(
f(3, 4, 9, 9)(32)
);

### How?

First we note that the sum of the T starting at n with height H can be broken down into two sums:

• n
• The sum of the T starting one row higher with height H - 1

By repeatedly adding n to the total, moving one row up, and subtracting 1 from H until it reaches 1, we end up summing the vertical stem of the T. Moving one row up is accomplished by subtracting x from n, since it can be observed that the difference between any cell and the one above is x.

When H reaches 1, we now have only the crossbar of width W left to sum. At this point n represents the center of the crossbar. We could sum the crossbar recursively as we did with the stem, but instead we take advantage of a fairly simple formula:

sum(n - a : n + a) = (n - a) + (n - (a-1)) + ... + n + ... + (n + (a-1)) + (n + a)
= (n - a) + (n + a) + (n - (a-1)) + (n + (a-1)) + ... + n
= 2n + 2n + ... + n
= n * (2a + 1)


In this case, our a is (W - 1) / 2, which makes

n * (2a + 1) = n * (((W - 1) / 2) * 2 + 1)
= n * ((W - 1) + 1)
= n * W


which is the sum of the crossbar.

• Seeing as you don't use y you just go with (x,w,h)=>Q=n=> and then call it using f(x,w,h)(n,y)? Seem to remember this coming up on Meta recently but can't remember the outcome. – Shaggy Sep 6 '17 at 20:32
• @Shaggy Thank you, found a better format for that trick :-) – ETHproductions Sep 6 '17 at 20:47

# C# (Visual C# Compiler), 64 52 bytes

(x,y,W,H,n)=>{for(y=0;H-->1;n-=x)y+=n;return y+W*n;}


Try it online!

The non-recursive answer did indeed turn out significantly shorter. Gross misuse of for loops and the fact that y is officially a mandatory input even though it's not used.

• Should return y+3*n; be return y+W*n? Otherwise you wouldn't need to pass in W. – Ayb4btu Sep 7 '17 at 0:59
• @Ayb4btu ... You know, I've actually had to fix that mistake multiple times along the way. I don't know how it keeps getting back in. – Kamil Drakari Sep 7 '17 at 2:27

# MATL, 12 bytes

:q*-t0)iq*hs


Inputs are: H, x, n, W.

Try it online!

### Explanation

:     % Implicit input: H. Push [1, 2, ..., H]
q     % Subtract 1. Gives [0, 1, ..., H-1]
*     % Implicit input: x. Multiply. Gives [0, x, ..., x*(H-1)]
-     % Implicit input: n. Subtract. Gives [n-0, n-x, ..., n-x*(H-1)]. This is
% the stem, bottom to top
t     % Duplicate
0)    % Get last element, that is, n-x*(H-1). This is the center top of the "T"
i     % Input: W
q     % Subtract 1
*     % Multiply. Gives n-x*(H-1)*(w-1). This is the sum of the vertical bar
% excluding its center, which is already accounted for in the stem
h     % Concatenate into a row vector
s     % Sum of vector. Implicit display


# Python 3, 38 bytes

lambda x,W,H,n:~-H*(x*(1-H/2-W)+n)+W*n


Try it online!

-4 bytes thanks to Jonathan Allan
-4 bytes thanks to Kevin Cruijssen/Ayb4btu

• EDIT: four bytes ...Save two bytes with our old friend tilde lambda x,W,H,n:n*H-x*H*~-H/2+~-W*(n-H*x+x) (FYI the TIO link has too many args in the call to f) – Jonathan Allan Sep 6 '17 at 21:25
• @JonathanAllan Oh whoops thanks. And yay thanks! :D – HyperNeutrino Sep 6 '17 at 21:29
• @LuisMendo Wait whoops I fixed that but forgot to update the link and then un-fixed it when golfing. Thanks! – HyperNeutrino Sep 7 '17 at 1:26
• lambda x,W,H,n:~-H*(x*(1-H/2-W)+n)+W*n is 4 bytes shorter. (Credit goes to @Ayb4btu's C# .NET answer.) – Kevin Cruijssen Sep 7 '17 at 13:20
• @KevinCruijssen Oh cool, thanks! – HyperNeutrino Sep 7 '17 at 13:39

# Perl 5, 52 bytes

($x,$y,$w,$h,$n)=<>;say$h*($n+$x*(.5-$h/2-$w))+$w*$n


Try it online!

• Do yoy need to take y as input, seeing as you're not using it? – Shaggy Sep 7 '17 at 8:45
• Not really, but it was part of the input spec, so I included it. – Xcali Sep 7 '17 at 13:09

# Jelly,  18 16  15 bytes

-1 byte thanks to Erik the Outgolfer (use of chain separator)

S’×⁵_ðc2Ḣ+⁸’P¤×


A full program taking [H,W], x, n in that order (you can add y to the arguments if you like, it's not used).

Try it online!

### How?

The total is:

• n multiplied by the number of squares used in the T (which is W+H-1)
• minus the width of the grid, x, times the triangle number of the height (1+2+3+...+h) to account for the lack as we go up the stem
• minus the width of the grid, x, times one less that the height (H-1) times one less that the width (W-1) to account for the lack at the top of the T, excluding the lack we already accounted for at the top of the stem.

That is:

(W + H - 1) * n - ((H - 1) * (W - 1) + Triangle(H)) * x


The triangle number of H is its binomial with 2 A.K.A. H-choose-2.

c2Ḣ+⁸’P¤× - Link 1: the x * (Triangle(H) + (H-1)*(W-1)): [H,W]; x
S’×⁵_ðc2Ḣ+⁸’P¤× - Main link: [H,W], x
S               - sum                                    H+W
’              - decrement                              H+W-1
⁵            - program's fifth argument (3rd input)   n
×             - multiply                               (H+W-1)*n
c2        - choose-2 (vectorises)                  [Triangle(H), Triangle(W)]
⁸    -    chain's left argument               [H,W]
’    -    decrement (vectorises)              [H-1,W-1]
P   -    product                             (H-1)*(W-1)
× - multiply (by chain's right arg, x)     ((H-1)*(W-1)+Triangle(H))*x
_           - subtract                               (H+W-1)*n-((H-1)*(W-1)+Triangle(H))*x


# Python 3, 38 bytes

lambda x,W,H,n:~-H*(n-x*(W-1+H/2))+n*W


Try it online!

HyperNeutrino's arithmetic expression with improved grouping.

# C# (.NET Core), 49 37 bytes

(x,y,W,H,n)=>~-H*(x*(1-H/2d-W)+n)+W*n


Try it online!

A direct formula avoiding the need for looping or recursion. Unfortunately I had to use 2d otherwise it was going to do integer division and truncate the fractional component. The y param is completely redundant and could be removed.

### Explanation

(n-x*(H-1))*W             // Calculates the Top of the T


(n-x*(H-1)) gets the cell index by subtracting from n the number of rows to go up multiplied by the grid width. Multiplying this by W gets the sum of the top bar of the T.

n*(H-1)-x*(H-2)*(H-1)/2   // Calculates the Stem of the T


I got this by using mathematical induction by trying to calculate the sum of 1, 10, 19, 28 where n=28, H=4, x=9. Which can be written as:

28 + (28-9) + (28-9-9) + (28-9-9-9)
28-9*0 + 28-9*1 + 28-9*2 + 28-9*3
28*4 -9*(0+1+2+3)
n*H -x*(1+2+3)
n*H -x*(3*(3+1)/2)
n*H -x*((H-1)*((H-1)+1)/2)
n*H -x*((H-1)*H/2)


but because we don't want to include the top cell of the stem (included in the formula for the Top of the T), H needs to be H-1. Making the formula

n*(H-1)-x*((H-2)*(H-1)/2)


Combining these two formulas gives

(n-x*(H-1))*W + n*(H-1)-x*((H-2)*(H-1)/2)


and simplifying it gives the formula used for the answer (though how it is rearranged can change how it looks).

### Acknowledgements

Saved 12 bytes thanks to Kevin Cruijssen. Though I'm not sure if the ; should be included or not.

• You can golf some things: =>{return ...} can be =>...; and (H-1) can be ~-H. So in total: (x,y,W,H,n)=>~-H*(x*(1-H/2d-W)+n)+W*n - 37 bytes – Kevin Cruijssen Sep 7 '17 at 13:03
• @KevinCruijssen Thanks, that's a huge saving, I'm not sure how I missed them... – Ayb4btu Sep 7 '17 at 19:25

# Pyth, 23 bytes

+-*KEQ**JEQctQ2*tE-K*Jt


Try it here! (note the order of the inputs)

# How?

+-*KEQ**JEQctQ2*tE-K*Jt    Full program. Q is input, E is evaluated input (reads a new line)

*KEQ                     The second input * the first input.
*JEQ                The third input * the first input.
ctQ2            Float division of the the first input decremented by 1, by 2.
*                    Product.
-                         Difference.
tE         The fourth input.
*Jt    The product of the third input and the first input decremented by 1...
-K        ... Subtracted from the second input
*           Product.
+                          Sum.
Output implicitly.


# Japt, 14 bytes

I tried a few different solutions based on ETH's observations but the shortest I've come up with (so far) is a straight port.

Takes input in the order n,H,W,x.

´V©ßXnU)+UªU*W


Try it

# Java (OpenJDK 8), 1009996 91 bytes

x->W->H->n->{int t=n,i=0,j=-W/2;for(;++i<H;t+=n-x*i);for(;j<W/2;t+=n-x*~-H+j++);return-~t;}


Try it online!

• When using two parameters currying is shorter; when using three parameters currying is the same length; but when using four or more parameters currying is actually longer. So you can replace x->W->H->n-> with (x,W,H,n)->, like this. – Kevin Cruijssen Sep 7 '17 at 14:44
• Also, you can save some additional bytes by changing t+=(n-x*(H-1))+j++ to t+=n-x*~-H+j++ so you don't need the four parenthesis. And return t+1; can be return-~t; so you don't need the space. Here is the relevant codegolfing tip post for using -~ and ~- with more information. – Kevin Cruijssen Sep 7 '17 at 14:46
• @KevinCruijssen Thank you, I'm just using currying because I don't know if the bytes of the functional interface are counted – Roberto Graham Sep 7 '17 at 14:53

# QBIC, 32 bytes

[0,:-1|i=:-(a*:)┘g=g+i}?i*:+g-i


This takes parameters in the order h, n, x, w, and y is ignored.

## Explanation

Each time a variable is read from the cmd line in QBIC, using the : command, it is assigned a letter (a-z). For clarity, I will use h, n, x, w in the explanation instead of the b,c,d,e they would be assigned (a gets taken by the FOR-loop).

First, we want to know the total of the stem
[0,h-1|     FOR a = 0 to the height - 1
i=n-(a*x)     helper i holds an increment value: this is n, minus the
grid-width once for every row we go up.
g=g+i         helper g is incremented by i for each row
}         NEXT
?         PRINT
i*w          the center on the top bar times the width (ie the average)
-i           corrected for that ione i we already counted in the stem
+g           plus the stem itself


# Java 8, 35 bytes

(x,W,H,n)->~-H*(x*(1-H/2d-W)+n)+W*n


Port of @Ayb4btu .Net C# answer, after I golfed it a bit.

Try it here.

If y is a mandatory input-parameter, it will be 37 bytes instead (Try it here):

(x,y,W,H,n)->~-H*(x*(1-H/2d-W)+n)+W*n


If outputting a double isn't allowed, it will be 40 bytes instead (Try it here):

(x,W,H,n)->~-H*(x*(int)(1-H/2d-W)+n)+W*n


If having a function instead of a full program isn't allowed, it will be 160 bytes instead (Try it here):

interface M{static void main(String[]a){Integer W=new Integer(a[2]),H=W.decode(a[3]),n=W.decode(a[4]);System.out.print(~-H*(W.decode(a[0])*(1-H/2d-W)+n)+W*n);}}