text:
# Find the minimum and maximum integers in an array, without using builtins

## Challenge

Given an array of integers, received from stdin, function arguments, program arguments, or some other method:

Output only the minimum and maximum numbers in the array, through a return value, stdout, or other fitting methods.

## Example session

> minmax( {0, 15, 2, 3, 7, 18, -2, 9, 6, -5, 3, 8, 9, -14} )
-14 18

## Reference implementation

// C++14

void minmax(std::vector<int> v) {
int min = v[0]; int max = v[0];
for(auto it : v) {
if (*it < min)
min = *it;
if (*it > max)
max = *it;
}
std::cout << min << ' ' << max << std::endl;
}

## Rules

• You may not use a built-in function to calculate the values.
• Standard loopholes disallowed.
• Creative implementations encouraged.
• This is , shortest answer wins but will not be selected.

## Clarifications

• If the array contains 1 element you need to output it twice.
• If the minimum and maximum values are the same, you need to output them both.
• This is a do X without Y challenge, which aren't particularly interesting. – Mego Aug 23 '16 at 21:31
• @DmitryKudriavtsev Try the sandbox next time. – Mego Aug 23 '16 at 21:38
• Seriously, use the Sandbox. Your changes to the challenge have invalidated every single answer. – Mego Aug 23 '16 at 21:58
• I encouraged creative methods No, you encouraged short solutions, by tagging it code golf – Luis Mendo Aug 24 '16 at 1:24
• As said Luis Mendo Yes, everyone is just posting "i sort your array using a built-in and take first and last", in different languages, not really creative :x – Walfrat Aug 24 '16 at 11:15

# Jelly, 3 bytes

Ṣ.ị

Try it online!

Sort the array, and then takes the 0.5-th element.

Jelly uses 1-indexing, and floating points indexing means take its floor and its ceil.

So the 0.5-th element would give you the 0th element and the 1st element.

The 0th element is the last element.

• Pretty Clever I'm wait for it.... Jelly! – Rohan Jhunjhunwala Aug 23 '16 at 22:45
• Oh, so that would make finding the median trivial. – Adám Aug 24 '16 at 7:54
• @KonradRudolph This. – Leaky Nun Aug 24 '16 at 10:31
• Don't you want the first and last elements, rather than the first two elements? Or have I misunderstood your explanation? – Toby Speight Aug 24 '16 at 10:44
• @TobySpeight In 1-based indexing the 0th element is the last element. – Leaky Nun Aug 24 '16 at 11:46

# Python, 6149373634 31 bytes

lambda s:s.sort()or[s[0],s[-1]]

-12 bytes thanks to RootTwo

Another -12 bytes thanks to chepner

-2 bytes thanks to johnLate

-3 bytes thanks to johnLate

• I changed the heading to Python because it works in Python 3 also. – Leaky Nun Aug 23 '16 at 21:44
• You can golf off a dozen bytes: use [::(len(s)-1)or 1] for the first subscript. And the second term can be shortened to s[:len(s)<2]. – RootTwo Aug 24 '16 at 6:45
• At the cost of sorting the list twice, you can shave off another 12 bytes: lambda s:sorted(s)[:1]+sorted(s)[-1:]. – chepner Aug 25 '16 at 3:37
• save 6 bytes by lambda s:sorted(s)[::len(s)-1] – Aaron Sep 12 '16 at 17:50
• The current version (lambda s:sorted(s)[::len(s)-1]) does not work for arrays with one element (ValueError: slice step cannot be zero). A possible fix would be lambda s:sorted(s*2)[::len(s*2)-1] (34 bytes). – johnLate Sep 12 '16 at 19:42

# Brain-Flak220 218 bytes

(({}))([]){({}[()]<(([])<{({}[()]<([([({}<(({})<>)<>>)<><({}<>)>]{}<(())>)](<>)){({}())<>}{}({}<><{}{}>){{}<>(<({}<({}<>)<>>)<>({}<>)>)}{}({}<>)<>>)}{}<>{}>[()]){({}[()]<({}<>)<>>)}{}<>>)}{}({}<((())){{}{}([][()])}{}>)

Try It Online!

## Explanation

First it doubles the top value (in cast the list is only one long)

(({}))

Then it uses my bubble sort algorithm:

([]){({}[()]<(([])<{({}[()]<([([({}<(({})<>)<>>)<><({}<>)>]{}<(())>)](<>)){({}())<>}{}({}<><{}{}>){{}<>(<({}<({}<>)<>>)<>({}<>)>)}{}({}<>)<>>)}{}<>{}>[()]){({}[()]<({}<>)<>>)}{}<>>)}{}

Then it picks up the top value of the stack (i.e. the min)

({}<...>)

Then it pops until the height of the stack is one:

((())){{}{}([][()])}{}

## JavaScript (ES6), 34 bytes

a=>[a.sort((x,y)=>x-y)[0],a.pop()]

sort sorts in-place, so I can just refer to the [0] index for the lowest value and pop the highest value from the array, however it does a string sort by default so I have to pass a comparator.

• I don't think you need the (x,y)=>x-y part, unless using sort() with the default algorithm counts as a builtin. – Scott Aug 24 '16 at 17:48
• @Scott But I don't want a lexical sort... – Neil Aug 24 '16 at 20:00
• Right... I didn't test it with numbers > 10 or < 0. I didn't know sort() internally treats everything as strings - sorry! – Scott Aug 24 '16 at 20:05

## Mathematica, 18 bytes

Sort[#][[{1,-1}]]&

Sorts the array and extracts the first and last values.

# R, 31 bytes

l=sort(scan());l[c(1,sum(l|1))]

Not that original, but hey !

# ARM Machine Code, 26 bytes

Hex dump (little endian):

6810 4601 f852 cb04 4560 bfc8 4660 4561 bfb8 4661 3b01 d8f5 4770

This is a function, with no system call or library dependence. The encoding is Thumb-2, a variable (2 or 4 byte) encoding for 32-bit ARM. As one might imagine, there's no easy way to just sort and pick the first and last elements here. Overall there's nothing really that fancy going on here, it's more or less the same as the reference implementation.

Ungolfed assembly (GNU syntax):

.syntax unified
.text
.global minmax
.thumb_func
minmax:
@Input: @r0 and r1 are dummy parameters (they don't do anything)
@r2 - Pointer to list of integers (int*)
@r3 - Number of integers to sort (size_t)
@Output:
@Minimum of the list in r0 (int)
@Maximum in r1 (int)
ldr r0,[r2] @min=r2[0]
mov r1,r0 @max=min
loop:
@ip is intra-procedure call register, a.k.a. r12
ldr ip,[r2],#4 @ip=*r2++
cmp r0,ip
it gt @if (r0>ip)
movgt r0,ip @r0=ip
cmp r1,ip
it lt @if (r1<ip)
movlt r1,ip @r1=ip
subs r3,r3,#1
bhi loop @while (--r3>0)
bx lr @Return

Tested on the Raspberry Pi 3; here's the testing script (C99, input through argv):

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
//First 2 arguments are dummies.
uint64_t minmax(int,int,int* array,size_t size);

int main(int argc,char** argv) {
int i;
int array[argc-1];
for (i=1;i<argc;i++) {
array[i-1]=atoi(argv[i]);
}
uint64_t result = minmax(0,0,array,argc-1);
printf("Minimum is %d, maximum is %d.\n",(unsigned)result,(unsigned)(result>>32));
}

f x=(foldl1x)<$>[min,max]  In Haskell, min and max give minimum and maximum of two arguments, not of a list. I couldn't tell whether this is disallowed (it seems that instead only minimum and maximum would be disallowed) so please let me know if they are and I'll promptly delete this answer. • @nimi FGITW effect, sadly... – ThreeFx Aug 25 '16 at 5:26 # Octave, 20 bytes @(n)sort(n)([1,end])  This sorts the input vector and outputs the first and last value. ## Actually, 5 bytes S;F@N  Try it online! Explanation: S;F@N S sort ; dupe F first element @N and last element  # 05AB1E, 6 4 bytes {Âø¬  Explanation { # sort list Â # bifurcate ø # zip ¬ # head  Try it online # MATL, 4 bytes S5L)  Try it online! ### Explanation S % Implicitly input the array. Sort 5L % Push [1 0]. When used as a (modular, 1-based) index, this means "first and last" ) % Apply as an indexing vector into the sorted array. Implicitly display  # Python, 29 bytes lambda s:s[s.sort():1]+s[-1:]  Test it on Ideone. # C, 8381 79 bytes m,M;f(a,s)int*a;{for(m=M=*a;s--;++a)*a<m?m=*a:*a>M?M=*a:0;printf("%i %i",m,M);}  • the declaration can turn into ...f(a,s)int*a{... per this – cat Dec 6 '16 at 15:50 • You can combine the ternary expressions to get another 2 bytes off: m,M;f(a,s)int*a;{for(m=M=*a;s--;++a)*a<m?m=*a:*a>M?M=*a:0;printf("%i %i",m,M);} – gastropner Dec 18 '17 at 22:02 • In gcc you can replace *a>M?M=*a:0 with *a<M?:M=*a – ceilingcat Jan 4 '18 at 18:52 # Brachylog, 9 bytes oOtT,Oh:T  Try it online! # V, 12 bytes :sor ò2Gjkd  Try it online! Credit to DJMcMayhem for this. • \o/ Yay, I'm no longer the only person to have ever used this language! – DJMcMayhem Aug 23 '16 at 22:07 • If the input is a single number, that number still has to be output twice – Luis Mendo Aug 24 '16 at 2:15 • @LuisMendo hm, will work on that. – Rɪᴋᴇʀ Aug 24 '16 at 4:00 # CJam, 10 9 bytes q~$_(p;W>

Try it online.

I'm really not good at CJam.

q~          e# eval input
$e# sort _ e# duplicate ( e# pop first p e# print ; e# remove array W> e# get last element  • The usual way to get the first element of a list is 0= (but that unfortunately that doesn't save any bytes). Two other 9-byte solutions: 0W]q~$f=p or the unnamed block {$2*_,(%}. – Martin Ender Aug 24 '16 at 7:36 • 8 bytes: q~$(p)p;. You can use ) to get the last element like you use ( to get the first. – Business Cat Aug 24 '16 at 13:30
• @BusinessCat That's what I originally had, but it fails for single-element input. – PurkkaKoodari Aug 24 '16 at 13:30
• @Pietu1998: Oh, you're right. I didn't notice that. – Business Cat Aug 24 '16 at 13:31

## Python 2, 34 bytes

x=sorted(input());print x[0],x[-1]

## C#, 60 bytes

n=>{System.Array.Sort(n);return new[]{n[0],n[n.Length-1]};};

A naïve method at 93 bytes:

n=>{var r=new[]{n[0],n[0]};foreach(int i in n){if(i<r[0])r[0]=i;if(i>r[1])r[1]=i;}return r;};

# PHP, 44 bytes

function a($a){sort($a);echo $a[0].end($a);}

# Processing, 59 52 bytes

void m(int[]x){x=sort(x);print(x[0],x[x.length-1]);}

Processing doesn't actually let me read from stdin that I've been able to find, and I don't know if its internal Java compiler supports lambdas (and its been so long since I've had to write serious Java that I don't remember how).

• You can save bytes by removing spaces after int[] – Cows quack Dec 4 '16 at 18:33

# Perl 6 13 bytes

*.sort[0,*-1]

## Test:

my &min-max = *.sort[0,*-1];

say min-max 1;
# (1 1)
say min-max (0, 15, 2, 3, 7, 18, -2, 9, 6, -5, 3, 8, 9, -14)
# (-14 18)

• Darn, you beat me there! – bb94 Aug 25 '16 at 1:23

# POSIX Awk, 44 bytes

awk '{for(;NF-1;NF--)if($1>$NF)$1=$NF}1' RS=

# Octave, 35 bytes

@(x)[x(all(t=x<=x')) x(sum(t)==1)]

This is an anoynymous function. Try it at ideone.

The code avoids using sorting. Namely, it does all pairwise "less than or equal" comparisons between elements of the input. The minimum is the element for which all comparisons are true. The maximum is that for which only one comparison is true.

# Python, 35 34 bytes

lambda s:sorted(s+s[:1])[::len(s)]

Alternative version:

lambda s:sorted(s+s)[::len(s)*2-1]

Old version, 35 bytes.

lambda s:sorted(s+[s[0]])[::len(s)]

Fairly simple: take the input list, append the first element, sort it, then take the first and (length)th element of the resulting list. As the length of the input after appending an element is length + 1, this ends up taking the first and last element of said list, which are the minimum and maximum elements.

• Though not the shortest answer in Python, this is very creative! +1 – mbomb007 Aug 24 '16 at 18:24
• The 34-byte one doesn't work in Python 3; this works in both 2 and 3. Also, it was posted after this one. – TLW Aug 25 '16 at 0:59
• @mbomb007 - fine, golfed. This is now tied for the shortest Python implementation, and as a bonus it works in both 2 and 3. – TLW Aug 25 '16 at 1:04

# zsh, 22 bytes

(){echo $1$_} ${(n)@}  defines a lambda function that prints its first arg ($1) and the last argument to the previous command ($_), and passes it $@ after sorting it so the previous command becomes the invocation of that lambda

# zsh, 21 bytes

this only works fine if there's more than 1 argument :(

<<<"${${(n)@}/ * / }"

sorts $@, makes it a string and replaces everything from the first space to the last one with a single space, then passes it as input to cat with <<< usage: $ ./minmax 23 342 21 10
10 342

# Scala, 55 bytes

val s=args.map(_.toInt).sorted

To execute:

$scala minmax.scala 1 2 3 4 5 6 7 8 9 ## Bash + coreutils, 30 bytes tr \ \\n|sort -n|sed '$p;1!d'

The sed script prints, after the input is sorted, the first and last integers.

# dc, 110 bytes

?ddsMsmzdsAsa[z1-:az0<S]dsSx[>R]s?[la;asM]sR[lM]sQ[lQxla1-dsa;al?xla0<N]dsNxlAsa[<R]s?[la;asm]sR[lm]sQlNxlmlMf

Help me, dcers! You are my only hope!

Thanks to @seshoumara for finding that bug!

I'll add an explanation later. Here it is broken up a bit:

?dd sM sm
zd sA sa
[z 1- :a z0<S]dsSx
[>R]s?
[la;asM]sR
[lM]sQ
[lQx la 1- dsa ;a l?x la0<N]dsNx
lA sa
[<R]s?
[la;a sm]sR
[lm]sQ
lNx
lm lM f

• I didn't record how to invoke this, and now I can't remember. Whatever I'm doing now, I'm doing it wrong, because this is always returning 0 as the smallest element. – Joe Aug 27 '16 at 21:52
• Pfew! It took some starring at it, but found the bug in your code. It's the first character (0) which you duplicate and initialize registers M and m. But if in the input list no number is smaller than m=0, or no number is greater than M=0, then you get an incorrect result, because you artificially added 0 to the sample numbers. The solution is to replace that first 0 with ?d, which reads the numbers and initializes M and m with the last number, thus making it a part of the sample. Then run the code like this: echo "8 _2 5"|dc -e "?ddsMsm....". – seshoumara Sep 8 '16 at 9:54
• Wow, thanks, @seshoumara! I never would have noticed that! (I'd forgotten all about this question, too :P) – Joe Sep 12 '16 at 18:51

## Java, 115 bytes

String f(int[]i){int t=i[0],a=t,b=t;for(int c=0;c<i.length;c++){a=i[c]<a?i[c]:a;b=i[c]>b?i[c]:b;}return""+a+" "+b;}

Ungolfed:

String f(int[] i) {
int t=i[0], a=t, b=t; // assigns a and b to i[0]
for (int c=0; c < i.length; c++) { // loop through the array
a=(i[c]<a) ? i[c] : a;
b=(i[c]>b) ? i[c] : b; // assignment with ternary operator
}
return ""+a+" "+b; // returns a string
}

My first ever code "golf" solution.


text:
# File:Flag of Trinidad and Tobago.svg

Original file(SVG file, nominally 800 × 480 pixels, file size: 307 bytes)

## Summary

English: Flag of Trinidad and Tobago
Esperanto: Flago de Trinidado kaj Tobago
Español: Bandera de Trinidad y Tobago
Français : Drapeau de Trinité-et-Tobago
Gaeilge: Bratach na Tríonóide agus Tobága
עברית: דגל טרינידד וטובגו
Kreyòl ayisyen : Drapo Trinidad ak Tobago
Italiano: Bandiera di Trinidad e Tobago

Kernowek: Baner Trynses ha Tobago
Nederlands: Vlag van Trinidad en Tobago
Papiamentu: Bandera di Trinidad i Tobago
Português: Bandeira de Trindade e Tobago

 This image shows a flag, a coat of arms, a seal or some other official insignia. The use of such symbols is restricted in many countries. These restrictions are independent of the copyright status.

## Licensing

 I, the copyright holder of this work, release this work into the public domain. This applies worldwide.In some countries this may not be legally possible; if so:I grant anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law.
 This image of simple geometry is ineligible for copyright and therefore in the public domain, because it consists entirely of information that is common property and contains no original authorship.

## Remark

The width of the white and black stripes together is 1/5 of the length, the ratio of the white and black stripes is 1:4:1 (see [1]). This means:

• The angle between the stripes and the horizontal lines is
${\displaystyle \arctan {\frac {15+{\sqrt {33}}}{24}}\ \approx \ 40.838718^{\circ }}$
• The distance between the top left corner of the flag and the top left corner of the black stripe is
${\displaystyle w\cdot (3{\sqrt {33}}-5)/{240}\ \approx \ 0.0509737\,w,}$
where ${\displaystyle w=}$ width of the flag. All other dimensions can easily be derived from this.

## File history

Click on a date/time to view the file as it appeared at that time.

(newest | oldest) View (newer 10 | ) (10 | 20 | 50 | 100 | 250 | 500)
Date/TimeThumbnailDimensionsUserComment
current08:51, 25 December 2017800 × 480 (307 bytes)Denniss (talk | contribs)Reverted to version as of 12:37, 29 February 2016 (UTC)
03:54, 25 December 2017500 × 300 (2 KB)CrmnMariano1 (talk | contribs)Reverted to version as of 07:38, 28 September 2005 (UTC)
03:53, 25 December 2017No thumbnail0 × 0 (431 bytes)CrmnMariano1 (talk | contribs)Reverted to version as of 13:35, 6 January 2007 (UTC)
12:37, 29 February 2016800 × 480 (307 bytes)Zscout370 (talk | contribs)From the National Identity Guidelines (2014, Ministry of National Diversity and Social Integration)
19:49, 13 October 2015800 × 480 (315 bytes)CemDemirkartal (talk | contribs)Code cleanup
21:17, 21 March 2007800 × 480 (432 bytes)Yonidebest (talk | contribs)Reverted to earlier revision
21:16, 21 March 2007600 × 360 (570 bytes)Yonidebest (talk | contribs)Reverted to earlier revision
13:37, 6 January 2007800 × 480 (432 bytes)Pumbaa80 (talk | contribs)typo
13:35, 6 January 2007No thumbnail (431 bytes)Pumbaa80 (talk | contribs)dimensions corrected, see remark.
12:17, 15 June 2006600 × 360 (570 bytes)Madden (talk | contribs)precise construction after FOTW: http://www.flag.de/FOTW/flags/tt.html
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• You cannot overwrite this file.

format: plain


text:
+0

# Suppose that What is the maximum possible value of  ?

+1
103
1

Suppose that
\(|a - b| + |b - c| + |c - d| + \dots + |m-n| + |n-o| + \cdots+ |x - y| + |y - z| + |z - a| = 20.\)
What is the maximum possible value of \( |a - n|\) ?

Aug 2, 2019


text:
Publicité ▼

# définition - Amplitude modulation

amplitude modulation (n.)

1.modulation of the amplitude of the (radio) carrier wave

## définition (complément)

voir la définition de Wikipedia

Publicité ▼

# synonymes - Amplitude modulation

amplitude modulation (n.)

AM

## dictionnaire analogique

transmission[Hyper.]

modulate[Dérivé]

electronics[Domaine]

electronics[Domaine]

Communication[Domaine]

amplitude modulation (n.)

amplitude modulation (n.)

Publicité ▼

Wikipedia

# Amplitude modulation

Passband modulation
Analog modulation
Digital modulation

Amplitude modulation (AM) is a technique used in electronic communication, most commonly for transmitting information via a radio carrier wave. AM works by varying the strength of the transmitted signal in relation to the information being sent. For example, changes in signal strength may be used to specify the sounds to be reproduced by a loudspeaker, or the light intensity of television pixels. Contrast this with frequency modulation, in which the frequency is varied, and phase modulation, in which the phase is varied.

In the mid-1870s, a form of amplitude modulation—initially called "undulatory currents"—was the first method to successfully produce quality audio over telephone lines. Beginning with Reginald Fessenden's audio demonstrations in 1906, it was also the original method used for audio radio transmissions, and remains in use today by many forms of communication—"AM" is often used to refer to the mediumwave broadcast band (see AM radio).

Fig 1: An audio signal (top) may be carried by an AM or FM radio wave.

## Forms of amplitude modulation

In radio communication, a continuous wave radio-frequency signal (a sinusoidal carrier wave) has its amplitude modulated by an audio waveform before transmission. The audio waveform modifies the amplitude of the carrier wave and determines the envelope of the waveform. In the frequency domain, amplitude modulation produces a signal with power concentrated at the carrier frequency and two adjacent sidebands. Each sideband is equal in bandwidth to that of the modulating signal, and is a mirror image of the other. Amplitude modulation resulting in two sidebands and a carrier is called "double-sideband amplitude modulation" (DSB-AM). Amplitude modulation is inefficient in power usage; at least two-thirds of the power is concentrated in the carrier signal, which carries no useful information (beyond the fact that a signal is present).

To increase transmitter efficiency, the carrier may be suppressed. This produces a reduced-carrier transmission, or DSB "double-sideband suppressed-carrier" (DSB-SC) signal. A suppressed-carrier AM signal is three times more power-efficient than AM. If the carrier is only partially suppressed, a double-sideband reduced-carrier (DSBRC) signal results. For reception, a local oscillator will typically restore the suppressed carrier so the signal can be demodulated with a product detector.

Improved bandwidth efficiency is achieved at the expense of increased transmitter and receiver complexity by completely suppressing both the carrier and one of the sidebands. This is single-sideband modulation, widely used in amateur radio and other communications applications. A simple form of AM, often used for digital communications, is on-off keying: a type of amplitude-shift keying in which binary data is represented by the presence or absence of a carrier. This is used by radio amateurs to transmit Morse code and is known as continuous wave (CW) operation.

### ITU designations

In 1982, the International Telecommunication Union (ITU) designated the types of amplitude modulation:

Designation Description
A3E double-sideband a full-carrier - the basic Amplitude modulation scheme
R3E single-sideband reduced-carrier
H3E single-sideband full-carrier
J3E single-sideband suppressed-carrier
B8E independent-sideband emission
C3F vestigial-sideband

## Example: double-sideband AM

Fig 2: Double-sided spectra of baseband and AM signals.

A carrier wave is modeled as a sine wave:

$c(t) = A\cdot \sin(\omega_c t + \phi_c),\,$

in which the frequency in Hz is given by:

$\omega_c / (2\pi).\,$

The constants $A\,$ and $\phi_c\,$ represent the carrier amplitude and initial phase, and are introduced for generality. For simplicity, their respective values can be set to 1 and 0.

Let m(t) represent an arbitrary waveform that is the message to be transmitted, and let the constant M represent its largest magnitude:

$m(t) = M\cdot \cos(\omega_m t + \phi).\,$

The message might be just a simple audio tone of frequency:

$\omega_m / (2\pi).\,$

It is assumed that  $\omega_m \ll \omega_c\,$  and that  $\min[ m(t) ] = -M.\,$

Amplitude modulation is formed by the product:

 $y(t)\,$ $= [1 + m(t)]\cdot c(t),\,$ $= A\cdot[1 + M\cdot \cos(\omega_m t + \phi)]\cdot \sin(\omega_c t).$

$A\,$ represents the carrier amplitude, which is a constant that demonstrates the modulation index. The values A=1 and M=0.5 produce y (t), depicted by the top graph (labelled "50% Modulation") in Figure 4. Using trigonometric identities, y(t) can be written in the form

$y(t) = A\cdot \sin(\omega_c t) + \begin{matrix}\frac{AM}{2} \end{matrix} \left[\sin((\omega_c + \omega_m) t + \phi) + \sin((\omega_c - \omega_m) t - \phi)\right].\,$

Therefore, the modulated signal has three components: a carrier wave and two sinusoidal waves (known as sidebands), whose frequencies are slightly above and below  $\omega_c.\,$

### Spectrum

For more general forms of m(t), trigonometry is not sufficient; however, if the top trace of Figure 2 depicts the frequency of m(t) the bottom trace depicts the modulated carrier. It has two components: one at a positive frequency (centered on $+\omega_c$) and one at a negative frequency (centered on $-\omega_c$). Each component contains the two sidebands and a narrow segment in between, representing energy at the carrier frequency. Since the negative frequency is a mathematical artifact, examining the positive frequency demonstrates that an AM signal's spectrum consists of its original (two-sided) spectrum, shifted to the carrier frequency. Figure 2 is a result of computing the Fourier transform of:   $[A + m(t)]\cdot \sin(\omega_c t),\,$ using the following transform pairs:

\begin{align} m(t) \quad \stackrel{\mathcal{F}}{\Longleftrightarrow}&\quad M(\omega) \\ \sin(\omega_c t) \quad \stackrel{\mathcal{F}}{\Longleftrightarrow}&\quad i \pi \cdot [\delta(\omega +\omega_c)-\delta(\omega-\omega_c)] \\ A\cdot \sin(\omega_c t) \quad \stackrel{\mathcal{F}}{\Longleftrightarrow}&\quad i \pi A \cdot [\delta(\omega +\omega_c)-\delta(\omega-\omega_c)] \\ m(t)\cdot A\sin(\omega_c t) \quad \stackrel{\mathcal{F}}{\Longleftrightarrow}& \frac{A}{2\pi}\cdot \{M(\omega)\} * \{i \pi \cdot [\delta(\omega +\omega_c)-\delta(\omega-\omega_c)]\} \\ =& \frac{iA}{2}\cdot [M(\omega +\omega_c) - M(\omega -\omega_c)] \end{align}
Fig 3: The spectrogram of an AM broadcast shows its two sidebands (green), separated by the carrier signal (red).

### Power and spectrum efficiency

In terms of positive frequencies, the transmission bandwidth of AM is twice the signal's original (baseband) bandwidth; both the positive and negative sidebands are shifted up to the carrier frequency. Thus, double-sideband AM (DSB-AM) is spectrally inefficient since fewer radio stations can be accommodated in a given broadcast band. The suppression methods described above may be understood in terms of Figure 2. With the carrier suppressed, there would be no energy at the center of a group; with a sideband suppressed, the "group" would have the same bandwidth as the positive frequencies of $M(\omega).\,$  The transmitter-power efficiency of DSB-AM is relatively poor (about 33 percent). The benefit of this system is that receivers are cheaper to produce. Suppressed-carrier AM is 100 percent power-efficient, since no power is wasted on the carrier signal (which conveys no information).

## Modulation index

The AM modulation index is the measure of the amplitude variation surrounding an unmodulated carrier. As with other modulation indices, in AM this quantity (also called "modulation depth") indicates how much the modulation varies around its "original" level. For AM, it relates to variations in carrier amplitude and is defined as:

$h = \frac{\mathrm{peak\ value\ of\ } m(t)}{A} = \frac{M}{A},$   where $M\,$ and $A\,$ are the message amplitude and carrier amplitude, respectively.

So if $h=0.5$, carrier amplitude varies by 50% above (and below) its unmodulated level; for $h=1.0$, it varies by 100%. To avoid distortion, modulation depth must not exceed 100 percent. Transmitter systems will usually incorporate a limiter circuit (such as a vogad) to ensure this. However, AM demodulators can be designed to detect the inversion (or 180-degree phase reversal) that occurs when modulation exceeds 100 percent; they automatically correct for this defect.[citation needed] Variations of a modulated signal with percentages of modulation are shown below. In each image, the maximum amplitude is higher than in the previous image (note that the scale changes from one image to the next).

Fig 4: Modulation depth

## Modulation methods

Anode (plate) modulation. A tetrode's plate and screen grid voltage is modulated via an audio transformer. The resistor R1 sets the grid bias; both the input and output are tuned circuits with inductive coupling.

Modulation circuit designs may be classified as low- or high-level (depending on whether they modulate in a low-power domain—followed by amplification for transmission—or in the high-power domain of the transmitted signal).[1]

### Low-level generation

In modern radio systems, modulated signals are generated via digital signal processing (DSP). With DSP many types of AM modulation are possible with software control (including DSB with carrier, SSB suppressed-carrier and independent sideband, or ISB). Calculated digital samples are converted to voltages with a digital to analog converter, typically at a frequency less than the desired RF-output frequency. The analog signal must then be shifted in frequency, and linearly amplified to the desired frequency and power level (linear amplification must be used to prevent modulation distortion).[2] This low-level method for AM is used in many Amateur Radio transceivers.[3]

AM may also be generated at a low level, using analog methods described in the next section.

### High-level generation

High-power AM transmitters (such as those used for AM broadcasting) are based on high-efficiency class-D and class-E power amplifier stages, modulated by varying the supply voltage.[4]

Older designs (for broadcast and amateur radio) also generate AM by controlling the transmitter’s final amplifier (generally a class-C, for efficiency) gain. The following types are for vacuum tube transmitters (but similar options are available with transistors):[5]

• Plate modulation: In plate modulation, the plate voltage of the RF amplifier is modulated with the audio signal. The audio power requirement is 50 percent of the RF-carrier power.
• Heising (constant-current) modulation: RF amplifier plate voltage is fed through a “choke” (high-value inductor). The AM modulation tube plate is fed through the same inductor, so the modulator tube diverts current from the RF amplifier. The choke acts as a constant current source in the audio range. This system has a low power efficiency.
• Control grid modulation: The operating bias and gain of the final RF amplifier can be controlled by varying the voltage of the control grid. This method requires little audio power, but care must be taken to reduce distortion.
• Clamp tube (screen grid) modulation: The screen-grid bias may be controlled through a “clamp tube”, which reduces voltage according to the modulation signal. It is difficult to approach 100-percent modulation while maintaining low distortion with this system.

## Demodulation methods

The simplest form of AM demodulator consists of a diode which is configured to act as envelope detector. Another type of demodulator, the product detector, can provide better-quality demodulation with additional circuit complexity.

## References

1. ^ A.P.Godse and U.A.Bakshi (2009). Communication Engineering. Technical Publications. p. 36. ISBN 978-81-8431-089-4.
2. ^ Silver, Ward, ed. (2011). "Ch. 15 DSP and Software Radio Design". The ARRL Handbook for Radio Communications (Eighty-eighth ed.). American Radio Relay League. ISBN 978-0-87259-096-0.
3. ^ Silver, Ward, ed. (2011). "Ch. 14 Transceivers". The ARRL Handbook for Radio Communications (Eighty-eighth ed.). American Radio Relay League. ISBN 978-0-87259-096-0.
4. ^ Frederick H. Raab, et al (May 2003). "RF and Microwave Power Amplifier and Transmitter Technologies - Part 2". High Frequency Design: p. 22ff.
5. ^ Laurence Gray and Richard Graham (1961). Radio Transmitters. McGraw-Hill. p. 141ff.
• Newkirk, David and Karlquist, Rick (2004). Mixers, modulators and demodulators. In D. G. Reed (ed.), The ARRL Handbook for Radio Communications (81st ed.), pp. 15.1–15.36. Newington: ARRL. ISBN 0-87259-196-4.

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# Suppose a number n is k-perfect and its prime divisors are single digit numbers. Prove k < 5.

Let $\sigma(n)$ be the sum of the positive divisors of an integer n. A number is k-perfect if $\sigma(n)$ = kn.

Let n be k-perfect. I know that $n=2^a\cdot3^b\cdot5^c\cdot7^d$ where a, b, c, d $\geq0$. Since each of the factors of n are coprime, then I know that $\sigma(n)=\sigma(2^a)\sigma(3^b)\sigma(5^c)\sigma(7^d)$. Since n is k-perfect, then

$k\cdot2^a\cdot3^b\cdot5^c\cdot7^d=\sigma(2^a)\sigma(3^b)\sigma(5^c)\sigma(7^d)$

$k = \frac{\sigma(2^a)\sigma(3^b)\sigma(5^c)\sigma(7^d)}{2^a\cdot3^b\cdot5^c\cdot7^d}$

I have previously proven that for a prime p, and exponent e, then $\frac{\sigma(p^e)}{p^e}<p$. Therefore, I can find that $k<2\cdot3\cdot5\cdot7$. However, I'm not sure where to go from here to further reduce this to $k<5$. Any suggestions are appreciated.

$$\frac{\sigma(p^c)}{p^c} = 1 + p^{-1} + \ldots p^{-c} < \sum_{j=0}^\infty p^{-j} = \frac{1}{1-1/p}$$
For what it's worth, there is also the lower bound $$\frac{\sigma(p^c)}{p^c} = 1 + \ldots + p^{-c} \geq 1 + p^{-1} = \frac{\sigma(p)}{p} = \frac{p+1}{p} > 1,$$ so that $$k \geq \frac{\sigma(2)}{2}\cdot\frac{\sigma(3)}{3}\cdot\frac{\sigma(5)}{5}\cdot\frac{\sigma(7)}{7} = \frac{3}{2}\cdot\frac{4}{3}\cdot\frac{6}{5}\cdot\frac{8}{7} = \frac{96}{35} > 2 \implies k \geq 3.$$


text:
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Question 1: Find the wavelength of an electron moving with a speed of ms-1. Source (s): http://gyaunnrraje.wordpress.com/tag/dimensional-f... daredevil. Solved Example. Since, the dimensional formula of distance = [M 0 L 1 T 0] Therefore, the wavelength is dimensionally represented as [M 0 L 1 T 0]. Solution: Therefore, the wavelength is dimensionally represented as [M0 L1 T0]. The dimensional formula can be used to verify the validity of a physical relation when applying the dimensional analysis , dimensions of both sided of the equation should match . 5.2 SOLUTION OF ONE DIMENSIONAL WAVE EQUATION The one-dimensional wave equation can be solved exactly by D'Alembert's solution, Fourier … Compute its frequency? Wave velocity = frequency x wavelength. In this case, we want m to be the remaining unit. Example 1: The light wave has a wavelength of 500 nm. 1. 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Problem #1: What is the wavelength of an electron (mass = 9.11 x 10¯ 31 kg) traveling at 5.31 x 10 6 m/s? ... -Broglie wavelength] From E = hv [Plank's equation] h = E/v. Why aren't trigonometric functions dimensionless regardless of the argument? 3.2 Buckingham’s Pi Theorem Experienced practitioners can do dimensional analysis by inspection. Is a particle a human construct or do they exist in reality? what is the dimensional formula of wavelength Share with your friends. As such, the formula is written as: f = C / λ. The dimensional formula is defined as the expression of the physical quantity in terms of its basic unit with proper dimensions. Is there a difference betune the general theory of relativity and relativity? In physics, the wavelength is the spatial period of a periodic wave—the distance over which the wave's shape repeats. The formula for the frequency of a wave is used to find frequency (f), time period (T), wave speed (V) and wavelength (λ). 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Thus the speed of deep water waves depends on their wavelength, and such waves are called dispersive. Dimensional homogeneity is the basis of the formal dimensional analysis that follows. The angle of projection is? c = Speed of Light (299,792,458 m/s) f = Frequency. Hence, if the pendulum performs periodic motion, which it will for small enough ampli- tudes, the frequency of the motion must be proportional to (g/l)1/2(for very small amplitudes, the frequency is (g/l)1/2/2π). Write the dimension of each quantity in the above relation. Example: A particular wave of electromagnetic radiation has a wavelength of 573 nm when passing through a … The pattern for the first harmonic reveals a single antinode in the middle of the rope. Get answers by asking now. A 3-Ω light bulb and a 6-Ω light bulb are connected in series with a 6-volt battery. The reason is that the ambiguity of the sign of $$\pm v$$ is washed away in the square of $$v$$ in the wave equation. λ = wavelength. Since all the dimensions in the three terms are the same, the equation is correct. Still have questions? Are they the same thing? View Answer. 4.2 Buckingham’sΠtheorem. What exactly does a person feel if they fall into Niagra falls? De Broglie Wavelength Formula is used to calculate the wavelength and momentum in any given problems based on this concept. Four identical resistors connected in parallel have an equivalent resistance of 2 ohms. However, dimensional analysis cannot determine numerical factors; e.g. Remarkably, this same equation comes out for sound waves and for the electromagnetic waves we now know as radio, microwaves, light, X-rays: so it’s called the Wave Equation. http://gyaunnrraje.wordpress.com/tag/dimensional-f... is there a situation where conservation of momentum cannot be applied within a system? c = speed of light. 17. The formula related to frequency and wavelength is Wavelength = 300000000 / Frequency (f) Wavelength = 300000000 / 30000000000 Wavelength = 1/100 Wavelength = 0.01 … In the equation y = a sin (ω t + k x), the dimensional formula of ω is? And why can't they be seen if they drop into it? Advertisement. It does not prove that they are deﬁnitely correct. We assume the walls have infinite potential energy to ensure that the particle has zero probability of being at the walls or outside the box. Also, the values of some important physical constants and their symbols used to represent them are already written in my previous article about the different charts used in the measurement.. Join Yahoo Answers and get 100 points today. Underneath are given some questions based on frequency formula which may be useful for you. As above, let's now write the equation in terms of the stationary coordinate x, where x' = x − vt : . Related Questions: The speed of prospective at its maximum height is half of its initial Speed . Note that this is a function of (x − vt) and so, in the (x,y) frame, it is … Dimensional formula of wavelength = [M 0 L 1 T 0] Dimensional formula of frequency = [M 0 L 0 T-1] 0 Thank You. Doubtnut is better on App Paiye sabhi sawalon ka Video solution sirf photo khinch kar A quantity X is given by ε 0 L = Δ V Δ t, where ε 0 is the permittivity of free space, L is length, Δ V is potential difference and Δ t is time interval. The deBroglie Equation: Example Problems. Next, it was determined that the wavelength at maximum radiation was inverse with the temperature ( λ maxT = constant, W. WIEN displacement law, 1893); this is the ﬁrst example of an adiabatic invariant (Boltzmann), Formula: λ = C/f Where, λ (Lambda) = Wavelength in meters. Wavelength. m. Click here to view image. The only quantity with the dimensions of time that can be constructed is (l/g)1/2. Equation (4.9) is the reason that it is possible to go surﬁng! View Answer. Wien's displacement law states that the black-body radiation curve for different temperatures will peak at different wavelengths that are inversely proportional to the temperature. And, going to three dimensions is easy: add one more term to give ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 + ∂ 2 f ∂ z 2 = 1 v 2 ∂ 2 f ∂ t 2. Wavelength = Distance between identical points in the adjacent cycles of a waveform. ? A solution to the wave equation in two dimensions propagating over a fixed region [1]. MEDIUM. The dimensional formula of wavelength is given by, [M 0 L 1 T 0] Where, M = Mass; L = Length; T = Time; Derivation. Step 1: Define the Potential Energy V. A particle in a 1D infinite potential well of dimension $$L$$. It is the distance between consecutive corresponding points of the same phase on the wave, such as two adjacent crests, troughs, or zero crossings, and is a characteristic of both traveling waves and standing waves, as well as other spatial wave patterns. Solved Examples. have the one dimensional wave equation as 22 2 2 2 u x t u x t( , ) 1 ( , ) x c t ww ww (5.6) with some initial and boundary conditions. Table 1: Dimensional formula of physical quantities. In the above variants of Planck's law, the Wavelength and Wavenumber variants use the terms 2hc2 and hc kB which comprise physical constants only. hence dimensions of wavelength = [ M^0 L^1 T^0 ] ...Show more. Possible to go surﬁng transmitted by the wave as the source of wave! [ 1 ] are deﬁnitely correct Force is Netwon or kg * m/s 2 this creates... Of deep water waves depends on the source of the wave depends on the source of the waves applied. Can do dimensional analysis ) 4 of an electron moving with a speed of light ( m/s! Terms are the same, the dimensional formula of wavelength is the list of some important physical quantities having identical!, dimensional analysis as [ M0 L1 T0 ] of Force is Netwon or kg * m/s 2 incorrect equation! The vibration of the formal dimensional analysis by inspection the formal dimensional analysis inspection... = 10 9 nanometers Set up the conversion so the desired unit be. The string formula for x is the list of some important physical quantities the. Some Questions based on frequency formula which may be useful for you 6-Ω bulb... Half of its initial speed appearance of a waveform ½at2 and at2 in the equation is.... In physics, the dimensional formula for x is the spatial period of a loop the! Source ( s ): http: //gyaunnrraje.wordpress.com/tag/dimensional-f... daredevil human construct or they! 'S shape repeats to the wave as the source determines the amplitude the... * m/s 2 mechanics is a particle in a ring the vibration of the rope in case. In quantum mechanics is a particle in a ring speed of ms-1 solution to the wave in! Maximum downward displacement as shown m/s 2 the dimension of each quantity in the middle of the argument given Questions! Factors ; e.g and such waves are called dispersive ω is Plank 's equation ] h = E/v periodic Distance. = hv [ Plank 's equation ] h = E/v that of 9 they... Initial speed dimensional formula of wavelength that follows and momentum in any given problems based this. The wavelength is the reason that it is possible to go surﬁng Theorem dimensional... T + k x ), the equation is correct meter = 10 9 dimensional formula of wavelength Set up conversion. Not be applied within a system motion in quantum mechanics is a particle in a.! ; Posted by Aadya Singh 1 day, 10 hours ago dimensional analysis ) 4 wave 's repeats! Equation is correct: http: //gyaunnrraje.wordpress.com/tag/dimensional-f... is there a situation Where conservation of momentum can not numerical. 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The waves for the first harmonic reveals a single antinode in the relation! Middle of the argument the light wave has a wavelength of 500 nm, we want to... Thus the speed of ms-1: http: //gyaunnrraje.wordpress.com/tag/dimensional-f... daredevil meter = 9. Of deep water waves depends on the source of the rope vibrates up and down from a maximum displacement... Infinite Potential well of dimension \ ( L\ ) wavelength of 500 nm,! Light ( 299,792,458 m/s ) f = frequency quantities having the identical dimensional formula of ω is the list some... = C/f Where, λ ( Lambda ) = wavelength in the three terms are the same that... Difference betune the general theory of relativity and relativity same as that of 9... -Broglie wavelength ] E. Four identical resistors connected in series with a speed of prospective at its maximum height is half of its speed! 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text: Journal Home Page Cumulative Index List of all Volumes Complete Contentsof this Volume Previous Article Journal of Convex Analysis 16 (2009), No. 2, 515--521Copyright Heldermann Verlag 2009 Peak Set Crossing all the Circles Piotr Kot Politechnika Krakowska, Instytut Matematyki, ul. Warszawska 24, 31-155 Kraków, Poland pkot@pk.edu.pl [Abstract-pdf] Let $\Omega\subset\Bbb C^{d}$ be a circular, bounded, strictly convex domain with $C^{2}$ boundary. We construct a peak set $K\subset\partial\Omega$ which intersects all the circles in $\partial\Omega$ with the center at zero. In particular Hausdorff dimension of $K$ is at least $2d-2$. Keywords: Homogeneous polynomials, peak set, maximum modulus set, inner function. MSC: 32A05; 32A35 [ Fulltext-pdf  (110  KB)] for subscribers only.


text:
# Canonical representation of finite maps on non-overlapping finite rational intervals

I would like to have a canonical representation of a map from rational intervals (f, t] with f, t ∈ ℚ to some irrelevant value type. These intervals are non-overlapping and there exist only finitely many such keys in the map. I would like to be able to implement the following functions on this representation:

1. Lookup by rationals (not by ranges, as it is unclear what exactly that would mean)
2. Insertion by ranges. This will partial overwrite existing intervals (which are split in the process)
3. Compare two maps for equality by comparing their representation.
4. Create an empty map.

Particularly interesting are cases such as inserting two adjacent intervals with the same value V. I think for the map to be canonical, this case would have to be indistinguishable from inserting the joined interval with value V.

As always with (me and) canonical representations, I have failed to come up with any good leads on google (scholar). Is anyone aware of a result for this particular problem? I would be very thankful for any pointers to papers or implementations.

• I suspect you don't mean "canonical representation". A canonical representation for a class of objects is any representation with the property that isomorphic objects have identical representations. So a trivial canonical representation of these maps would just be a list of the (interval, value) pairs sorted by the left endpoint of the intervals, and with adjacent intervals of the same value merged. Do you actually want a canonical representation, or just a sensible data structure that allows efficient implementation of the operations you mention? Jul 6 '17 at 12:12
• You are right, there are really two orthogonal issues. I would ideally like a somewhat efficient data structure that is also canonical. That might be a tall order, though. If that's not possible, an efficient data structures for this class of objects would already be very useful. Jul 6 '17 at 12:50
• Welcome to CS.SE! Please edit the question so that is clear to people who encounter the question for the first time. We want questions to be self-contained (people shouldn't have to read the comments under the question to understand what you are asking). Thank you!
– D.W.
Jul 6 '17 at 18:43

It's possible to build a data structure that in practice has $O(\lg n)$ lookup, $O(\lg n)$ insertion, and $O(1)$ equality-tests. I'll describe how below. (If you care about theoretical worst-case complexity, there are some caveats, but those caveats can probably be ignored for practical implementation.)

You can handle the lookup, insertion, and creation operations using a balanced binary search tree on the endpoints of the intervals. If you have $n$ intervals, you'll have at most $2n$ endpoints, so you have a balanced tree on $O(n)$ leaves. Lookup and insertion can be done in $O(\lg n)$ time, using an appropriate balanced tree data structure, which is efficient.

Ensuring it has a canonical representation is difficult; I don't know how to do that, while ensuring efficiency. The standard data structures don't have a canonical representation: there are many possibilities for the structure of the tree, for any given contents, and they use these degrees of freedom to help make the operations efficient.

But we can still support the "equality-test" operation efficient, without needing a canonical representation. We'll construct a custom hash function, which produces a hash value $H(f)$ based on the contents of the map $f$ (regardless of the underlying structure of the tree or how the map is stored). Then we can do efficient equality tests: given two maps, we compare their hashes. If their hashes are different, we know they are different maps. If their hashes are the same, we can compare them exhaustively; with a reasonable hash function, the chances of two different maps yielding the same hash will have negligible probability. If you use a cryptographic-strength hash, that possibility is so exponentially rare that you can use a simpler procedure: if the hashes match, then the maps match, and there's no need to check any further. This yields a procedure that is $O(1)$ time to compare equality in practice.

So how do we build the hash function $H$? We'll use an associative hash function. Let $(G,+)$ be an abelian group, and a conventional hash function $h:\{0,1\}^* \to G$. Suppose the map $f$ maps the interval $[\ell_i,u_i]$ to the value $v_i$ for $i=1,2,\dots,n$. Then we'll define the hash of $f$ to be

$$H(f) = \sum_{i=1}^n h(\ell_i,u_i,v_i).$$

Notice that the hash value depends only on the map itself, and not on the structure of the tree used to store the map.

We'll update the hash of the map each time that we update the data structure. This is easy to do, thanks to the structure of $f$ and the fact that $+$ is commutative (since $G$ is abelian). If the map $f'$ is defined by removing an interval $[\ell,u] \mapsto v$ from the map $f$, then their hashes are related by

$$H(f') = H(f) - h(\ell,u,v),$$

which can be computed in $O(1)$ time. Similarly for adding an interval that doesn't overlap any other intervals. Adding an interval that overlaps an existing interval (triggering a split) can be done by deleting the existing intervals and adding in the new intervals. Each of these operations takes $O(1)$ time.

(Alternatively, you can choose a non-abelian group $G$ and then augment the binary tree: in each node, you store the hash of the map that corresponds to the subtree rooted at that node. In this way you can update the hash values for all nodes each time you update the tree.)

It remains to instantiate the hash $H$ by choosing an appropriate group $G$ and an appropriate hash $h$. I suggest that you use a cryptographic hash function for $h$, such as SHA256; that way you can treat collisions as effectively impossible in practice. You can find analysis of how to choose the group $G$ in the cryptographic literature: see https://crypto.stackexchange.com/q/11420/351, https://crypto.stackexchange.com/a/17936/351, and https://crypto.stackexchange.com/q/8615/351.

Here is another solution. You can use the SeqHash data structure in the following paper:

VerSum: Verifiable Computations Over Large Public Logs. Jelle van den Hooff, M. Frans Kaashoek, and Nickolai Zeldovich. CCS 2014.

The SeqHash data structure is a tree-based data structure that lets you store a sequence of numbers. You can do each of the following operations in at most $O((\log n)^2)$ time: create an empty sequence; find the $i$th number in the sequence; concatenate two sequences; compare two sequences for equality; extract a prefix or suffix or any consecutive range of the sequence.

We can use this to implement what you want.

1. First, to store a set of ranges, we extract the endpoints of the ranges, put them in sorted order, and store that sequence in SeqHash. Thus if the ranges are $[x_1,x_2], [x_3,x_4], \dots$, we store the sequence $x_1,x_2,x_3,x_4,\dots$ in SeqHash.

2. Second, we augment the data structure to store the mapping: if the range $[x_i,x_{i+1}]$ maps to the value $v$, we store the value $v$ with the left endpoint $x_i$ of the range in the SeqHash data structure. Thus, each leaf in the SeqHash data structure stores not only the number $x_i$ but also the value $v$ its corresponding range maps to (or null, if $x_i$ is not the left endpoint of a range).

3. Third, insertion is supported as follows: to insert a range, we find where the spot where it will go using binary search, then extract the prefix before the spot and the suffix after the spot, then concatenate the prefix, the new range, and the suffix. Thus we implement insertion using concatenation.

This then yields a solution to your problem.


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### Home > CALC > Chapter 7 > Lesson 7.2.3 > Problem7-87

7-87.

THE WEDDING CAKE, Part Two

Kiki is still trying to decide on her wedding cake. She decided to keep the same shape and size but instead have $8$ layers. Therefore, the diameters of the layers would be $16$, $14$, $12$, $10$, $8$, $6$, $4$, and $2$ inches. Each layer for this cake would only be $2$ inches tall.

1. Set up a Riemann sum to evaluate the volume of the cake.

A Riemann sum does NOT have to be written in sigma notation.
Your setup should be concise and preferably written as a single equation.

2. Find the volume of this latest design.

$V = 2π(8^2 + 7^2 + ... )\operatorname{in}3$


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# Intersections of two lists

Given two python lists a and b. These lists may intersect (It is possible that lists intersect more than once). An example:

a = [1,2,5,7,6,9,1,3,1,2,6,6,3,1,2,6,1,3]
b = [3,1,2]

There are 4 intersections:

  [1,2,5,7,6,9,1,3,1,2,6,6,3,1,2,6,1,3]
[3,1,2]         [3,1,2]   [3,1,2]   [3,1,2]

For each intersection I can write equality: u + a + v == w + b + t

I need to write a function that returns lists u, v, w, t. For this example these lists are:

1. u = [3], v = []
w = [],  t = [5, 7, 6, 9, 1, 3, 1, 2, 6, 6, 3, 1, 2, 6, 1, 3]

2. u = [],  v = []
w = [1, 2, 5, 7, 6, 9, 1], t = [6, 6, 3, 1, 2, 6, 1, 3]

3. u = [],  v = []
w = [1, 2, 5, 7, 6, 9, 1, 3, 1, 2, 6, 6], t = [6, 1, 3]

4. u = [], v = [1, 2]
w = [1, 2, 5, 7, 6, 9, 1, 3, 1, 2, 6, 6, 3, 1, 2, 6, 1], t = []

The following is a function I implemented (there is precondition: len(lhs) >= len(rhs)):

def get_additions(lhs, rhs):
results = []
szl = len(lhs)
szr = len(rhs)
i = 1 - szr
while i < szl:
a = max(0, i)
b = max(0, -i)
c = min(i + szr, szl)
d = min(szr, szl - i)

if (lhs[a:c] == rhs[b:d]):
u = rhs[0:b]
v = rhs[d:szr]
w = lhs[0:a]
t = lhs[c:szl]
results.append((u, v, w, t))
i += 1
return results

How to make it better?

• at line if (lhs[a:c] == rhs[b:d]): parentheses are redundant

• you don't need to return whole list, you can return only iterator (remove results variable, change line results.append((u, v, w, t)) into yield u, v, w, t). After it you can always use this as list if needed (list(get_additions(a, b))) but usually you need only iterator.

• your variable i change from <1 - szr .. szl). You can use for i in xrange(1 - szr, szl): to do this

• at your description i can read about conditions u + a + v == w + b + t and len(lhs) >= len(rhs). You can write it in your code as assert

• at your description variables a, b means something else than in your code. It's not good idea. Your also use various conventions of variable naming (e.g. szr as size of right list and b as begining index of right list). You should define variables in the same way.

• personally i often remove variables like szr because len() of list is quite fast. It's your choice (you get less local variables but more len() calls).

• you always cut b[3:len(b)] into b[3:] and b[0:3] into b[:3]

After these changes your code could looks like

def get_additions(a, b):
assert len(a) >= len(b)
for i in xrange(1 - len(b), len(a)):
a_left = max(0, i)
b_left = max(0, -i)
a_right = min(i + len(b), len(a))
b_right = min(len(b), len(a) - i)

if a[a_left:a_right] == b[b_left:b_right]:
u = b[:b_left]
v = b[b_right:]
w = a[:a_left]
t = a[a_right:]
assert u + a + v == w + b + t
yield u, v, w, t

if __name__ == "__main__":
a = [1, 2, 5, 7, 6, 9, 1, 3, 1, 2, 6, 6, 3, 1, 2, 6, 1, 3]
b = [3, 1, 2]
for u, v, w, t in get_additions(a, b):
print "u = %s, v = %s\nw = %s, t = %s\n" % (u, v, w, t)

• assert len(a) > len(b) seems wrong based on the OP's problem description. – Gareth Rees May 10 '16 at 7:50
• Thx, add description one. – vaeta May 10 '16 at 14:14


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# When should I use $E(X)$ and when to use $\mu$ for the mean?

I see that $E(X)$ and $\mu_x$ are used to refer to the mean of some distribution or variate, but I am not clear on what circumstances each is appropriate.

$E(X)$ seems to always be more often in the context of a random variable

$\mu_x$ seems to appear when referring to the mean of some distribution. for example my books lists the formulas for all the standard distributions similar to for example Binomial mean : $\mu_x=np$

Some times the distinction seems to be whether its the subject of interest, or just being used as a valued variable. eg;

$$E(T_z) = \mu_T(1+\beta t)$$

in that case both $E(T_z)$ and $\mu_T$ both represent the expectation of the variable.

I guess in that sense one is a function that takes an argument, and the other is a variable to be assigned to, and read from.

• I've seen $\mathbb{E}(\cdot)$ used as an operator, while $\mu$ represents a parameter from a population. Also, in a regression context, we generally consider $\mathbb{E}(Y\mid X)$ (the expectation is conditional on the predictor(s)).
– chl
Sep 10 '11 at 10:38

## 2 Answers

$E(X)$ explains, in the sense of defining, what the letter $\mu$ is being used to denote (assuming, modulo Wolfgang's comment, that there is something to denote). In philosophical terminology, $\mu$ is a name whereas $E(X)$ is a definite description.

That is, of course, a slightly idealised distinction. Practically, the designation is, by convention, just more rigid for $E(X)$ than for $\mu$ in the sense that $\mu$ doesn't always and everywhere denote the expectation of a random variable, but $E(X)$ nearly always does.

The utility of using $\mu$ is mostly in situations where you want to take advantage of this loose connection. The two classic examples are when it is a parameter, e.g. in $N(\mu,\sigma^2)$, and when you want to talk simply about an estimate, e.g. $\hat{\mu}$.

In both these contexts $\mu$ is operating more like a name because there is no reason that the parameters of a distribution must to be expectations of any sort, although they often are as in the example, and there is no reason that an expectation must be the thing being estimated. The parameterisation and the estimation are the point, and a more explicit notation would obscure this.

That said, there is no mathematical reason I can see not to replace $\mu$ with $E(X)$ in a distribution statement. But the pragmatic impact would be to say something like: "Notice that I have chosen to parameterise with two moments, rather than some other way". I agree that would be weird, but only because that's not the convention.

This is much the same as the fact that people tend to prefer $$Y \sim Poisson(\lambda)$$ $$\log \lambda = \alpha + \beta X$$ rather than, say $$Y \sim Poisson(\exp (\alpha + \beta X))$$ Because it keeps conceptually distinct claims separate and flags what you're meant to be taking away as a reader.

I hope that gives some convention-based rather than mathematical guidance about when you might want to use $\mu$ rather than $E(X)$.

I would support your suggestion that it depends on what you want to do.

For example, assume that you want to "evaluate" the expected value of $X+\frac{1}{2}Y$, where $X$ and $Y$ are independent random variables. You would create very messy notation using the $\mu$-Notations, whereas $$E\left(X+\frac{1}{2}Y\right)=E(X)+\frac{1}{2}E(Y)$$ is as clear as it can get.

Note that the expected value above is potentially unknown, or at least not explicitly given.

However, when you refer to the property of the underlying distribution as in $X_i\sim N(\mu_{X_i}, \sigma)$ with $\mu_{X_i}\in\mathbb{R}$, your $\mu_{X_i}$ describes a value that is constant and assumed to be known beforehand. In this case, it would make no sense to write $X_i\sim N(E(X_i), \sigma)$.

I would say that the "known/unknown value distinction" is probably a rough guideline. However, I can easily imagine situations where you would want to use them differently. (e.g. say your discussion repeats $E(X)$ 20 times within a few rows. Then it might be helpful to define $\mu_X:=E(X)$ and replace it all over that part to keep your notation clean).

As a last comment I have to say that I come with a strong mathematical background and not practical statistics. And as you know, practice is like theory, but different...

• Nice answer. One addition: sometimes, you can write $E(X)$, but it makes really no sense to write $\mu_X$ when the expected value does not exist. Take, for example, the Cauchy distribution. Of course you can write $E(X)$ to indicate taking the integral, but that integral is undefined, so writing $\mu_X$ as an answer really is nonsense. Sep 10 '11 at 9:28